Start with the derivative of several functions such as polynomial, rational, exponential, logarithmic and trigonometric functions. A critical point, maximum, minimum, second derivative test and approximation are discussed.

Chapter 9

Application of Differentiation

We have studied the derivative of functions such as polynomial, rational, exponential, logarithmic and trigonometric functions. We now are ready to apply these derivatives in:
  • finding the equations of tangent and normal lines,
  • finding the critical points to analyze the maximum, minimum and inflection points,
  • testing the concavity,
  • finding the maxima and minima,
  • solving the approximation problems.


    • 9.1 Tangent Line and Normal Line

    We first introduce the inclination of the line and its slope. The inclination of a line that intersects the x-axis is the measure of the smallest nonnegative angle which the line makes with the positive end of the x-axis. We shall use the symbol θ to represent the angle of inclination.
    We here discuss the expression related to the inclination, namely, the slope of a line, say m, is the tangent of the inclination,

    m = tan θ .
    We see that the vertical lines have inclination 90° but no slope. A horizontal line has a slope and that slope is the number 0.

    If θ is as shown in either of the two positions in the following figure, then
    tan θ = y⁄x .
    y x 0 θ x P(x,y) y
    y x 0 θ x P(x,y) y
    Suppose we have a line with a pair of points, P1 = (x1, y1) and P2 = (x2, y2), on it.
    View post on imgur.com

    From the figure, we can write as
    m = y2 - y1⁄x2 - x1 = tan φ
    Since θ = φ, we get
    m = tan θ.

    View post on imgur.com

    From the figure, we can write as
    m = y1 - y2⁄x2 - x1 = - y2 - y1⁄x2 - x1 = - tan θ .

    Next we recall the notion of equations of tangent and normal line.
    For the curve y = f(x), the gradient of the tangent ℓ1 at the point (x1, y1) is the value of - y'(x) at x = x1, hence the equation of the tangent at (x1, y1) is
    y - y1 = y' (x1)(x - x1) .
    The line ℓ2 which is perpendicular to the tangent ℓ1 at (x1, y1) is called the normal to the curve at (x1, y1). Hence its gradient is the value of -(y' (x1) )-1 where y'(x1) ≠ 0 and the equation of the normal at (x1, y1) is
    y - y1 = - 1 ⁄ y'(x1) (x - x1).
    View post on imgur.com

    Example 1.
    Find the equations of tangent and normal line to the curve y = ln x at the point y = -1.
    Solution
    y = ln x,
    y' = 1⁄x .
    When y = -1, ln x = -1,
    x = e-1 = 1⁄e .
    At the point (1⁄e , -1), the gradient of tangent is y' (1⁄e) = e, and the gradient of normal is 1⁄e. The equation of tangent at (1⁄e , -1) is
    y - (-1) = e(x - 1⁄e)
    y = ex - 2.
    The equation of normal at (1⁄e , -1) is
    y - (-1) = - 1⁄e (x - 1⁄e)
    y = - 1⁄ex + 1⁄e2 - 1.

    Example 2.
    Find the equation of tangent to y = sin x at the origin.
    Solution
    y = sin x,
    y' = cos x.
    At (0,0), the gradient of tangent is y' (0) = cos 0 = 1.
    The equation of tangent is
    y - 0 = 1(x - 0)
    y = x.

    Example 3.
    Consider the curve f(x) = x2 + ax + b where a and b are constants. The tangent to this curve at the point x = 2 is y = 2x + 1. Find the values of a and b.
    Solution
    f(x) = x2 + ax + b,
    f'(x) = 2x + a.
    At x = 2, f(2) = 2a + b + 4 and f'(2) = a + 4.
    By problem, the gradient of tangent line is 2, so a + 4 = 2 which gives a = -2.
    The tangent line equation at (2, 2a + b + 4) is
    y - (2a + b + 4) = (a + 4) (x - 2).
    Substitute the value a = -2 in above equation, then
    y - b = 2(x - 2)
    y = 2x + b - 4.
    Since the tangent line equation at x = 2 is y = 2x + 1, it follows that
    b = 5
    Therefore a = -2 and b = 5.

    Exercise 9.1

    1. Find the equations of tangent and normal line to the curve y = 2 ln x at (1, 0).

    2. Find the equation of the line which passes through the origin and tangent to y = ex.

    3. Find the points of contact where horizontal tangent meet the curve y = x4 - 2x2 + 2.

    4. Consider the curve y = a √ x   + b ⁄ √ x where a and b are constants. The normal to this curve at the point at x = 4 is 4x + y = 22. Find the values of a and b.

    9.2 Maximum and Minimum

    Before we describe the critical pionts let us discuss the behavior of increasing and decreasing functions.

    Increasing and Decreasing Functions

    Consider the function on an open interval.

    Definition
    A function f is increasing on an interval (a, b) if for any two numbers x1 and x2 in (a, b),
    x2 > x1 implies f(x2) > f(x1),
    namely, an increase in x produces an increase in y.
    View post on imgur.com

    A function f is decreasing on an interval (a, b) if for any two numbers x1 and x2 in (a,b),
    x2 > x1   implies   f(x2) < f(x1),
    namely, an increase in x produces a decrease in y.
    View post on imgur.com"

    We can determine whether a function is increasing or decreasing at a given interval by looking into its graph. Sometimes we can get the same information by using the first derivative.
    Test for Increasing and Decreasing Functions
    Let y = f(x) be differentiable on the open interval (a, b).
    1. If f'(x) > 0 for each x ∈ (a, b), then f is increasing on (a, b).
    2. If f'(x) < 0 for each x ∈ (a, b), then f is decreasing on (a, b).

    Example 4.
    Show that the function f(x) = x2 + 1 is decreasing on the interval x < 0 and increasing on the interval x > 0.
    Solution
    f(x) = x2 + 1,
    f'(x) = 2x.
    On x < 0, f'(x) = 2x < 0. Thus f is decreasing on x < 0
    On x > 0, f'(x) = 2x > 0. Thus f is increasing on x > 0

    Example 5.
    Consider the function f(x) = x3 - 3x + 1.
    (a) Find the open intervals on which f is increasing and those on which f is decreasing.
    (b) Find the points at which the graph has a horizontal tangent.
    Solution
    (a) f(x) = x3 - 3x + 1.
    f'(x) = 3x2 - 3
    = 3(x - 1)(x + 1).
    So f'(x) > 0, when x - 1 and x + 1 have the same sign. Similarly, f'(x) < 0, when x - 1 and x + 1 have the opposite signs.

    Interval x < -1 -1 < x < 1 x > 1
    x + 1 - + +
    x - 1 - - +
    sign of f'(x) + - +
    Behavior of f   increasing   decreasing   increasing  

    Thus f(x) is increasing on x < -1 and x > 1, and decreasing on -1 < x < 1.
    (b) The graph has a horizontal tangent when f'(x) = 0.
    3(x - 1)(x + 1) = 0.
    x = 1 or x = -1.
    So f(-1) = 3, f(1) = -1.
    The horizontal tangents are at the points (-1, 3) and (1, -1).

    Critical Point

    We now describe how to decide the maximum and minimum points of the function by determining the values of its derivatives. It is one of the most important application of differentiation. We first introduce the notion of critical point.

    Definition
    A critical point of a function f(x) is a point in the domain of the function such that either f'(x) = 0 or f'(x) is undefined.
    f '(x) = 0
    f '(x) = 0
    f '(x) = 0
    f '(x) is undefined
    f '(x) is undefined
    Test for Increasing and decreasing function on an Interval
    Let f(x) be a function.
    Step 1 Find f '(x).
    Step 2 Find the critical points of f.
    Step 3 Find the open intervals determining by the critical points.
    Step 4 Test the signs of f '(x) for each interval.
    Step 5 Decide whether f is increasing or decreasing on each interval.

    Example 6
    Find the critical points of:
    (a) f(x) = x3 - 3x + 2
    (b) f(x) = (x - 1) 2⁄ 3
    (c) f(x) = 1⁄x
    Solution
    (a) curve: f(x) = x3 - 3x + 2
    f '(x) = 3x2 - 3.
    For critical points, f ' (x) = 0 so
    3x2 - 3 = 0.
    Therefore x = ± 1.
    When x = 1,
    f '(1) = 13 - 3(1) + 2 = 0.
    When x = -1,
    f(-1) = (-1)3 - 3(-1) + 2 = 4.
    Therefore the critical points are (1, 0) and (-1, 4).

    (b) curve: f(x) = (x - 1) 2⁄ 3
    f '(x) = 2⁄3(x - 1)- 1⁄3
    = 2⁄3(x - 1)1⁄3 .
    It is observed that f ' (x) is undefined when x = 1. Therefore the critical point is (1, 0).

    (c) curve: f(x) = 1⁄x
    f ' (x) = - 1⁄x2.
    It is observed that f ' (x) is undefined when x = 0.
    However, x = 0 is not in the domain of the function.
    Therefore, there are no critical points.

    Example 7.
    Find the open intervals on which the following functions are increasing or decreasing.
    (a) f(x) = 2x3 - 6x + 5
    (b) f(x) = (x2 - 9) 4⁄ 5
    (c) f(x) = 2x - 3⁄x - 1

    Solution
    curve: f(x) = 2x3 - 6x + 5
    f ' (x) = 6(x2 - 1) = 6(x + 1)(x - 1).
    For critical points, f ' (x) = 0, namely x = ±1.

    Interval x < -1 -1 < x < 1 x > 1
    x + 1 - + +
    x - 1 - - +
    sign of f '(x) + - +
    Behavior of f   increasing   decreasing   increasing 

    Therefore, on x < -1 and x > 1, the given function is increasing, and on -1 < x < 1, the given function is decreasing.

    (b) f(x) = (x2 - 9) 4⁄ 5
    curve: f(x) = (x2 - 9) 4⁄ 5
    f '(x) = 8⁄5 x⁄(x2 - 9)1⁄5
    For critical points, f '(x) = 0, namely x = 0.
    f '(x) is undefined, namely x = -3 and x = 3.

    Interval x < -3 -3 < x < 0 0 < x < 3 x > 3
    x + 3 - + + +
    x - - + +
    x - 3 - - - +
    sign of f '(x) - + - +
    Behavior of f   decreasing   increasing   decreasing   increasing  

    Therefore, on x < -3 and 0 < x < 3, the given function is decreasing, and on -3 < x < 0 and x > 3the given function is increasing.

    (c) f(x) = 2x - 3⁄x - 1
    curve: f(x) = 2x - 3⁄x - 1
    f ' (x) = 1⁄(x - 1)2.
    It is observed that x = 1 does not belong to the domain of function.

    Interval x < 1 x > 1
    (x - 1)2 + +
    sign of f ' (x) + +
    Behavior of f   increasing   increasing  

    Therefore, on both x < 1 and x > 1, the function is increasing.

    First Derivative Test

    Consider the following graph which has a restricted domain of -5 ≤ x ≤ 5.
    View post on imgur.com

    In the figure, points A and C are minimum, and points B and D are maximum. The point A is called a global minimum because f (x) has a minimum value on the entire domain. Similarly the point D is called a global maximum because f (x) has a maximum value on the entire domain. B is a local maximum because it is a turning point on the graph and f (x) has a maximum value at B compared to the neighboring points. C is a local minimum because it is a turning point on the graph and f (x) has a minimum value at C compared to the neighboring points.
    Test for Global Maximum and Global Minimum on a Closed Interval
    Let f (x) be a continuous function on a closed interval.
    Step 1 Find the critical points of f.
    Step 2 Find f (x) at all critical points and endpoints of the interval.
    Step 3 Take the smallest value as the global minimum and the largest value as the global maximum.

    Example 8.
    Find the global minimum and global maximum values of each of the following function:
    (a) f (x) = x2 + 5x - 3     on [-4, 5],
    (b) f (x) = x⁄x2 + 1     on [-3, 3].
    Solution
    (a) curve: f (x) = x2 + 5x - 3     on [-4, 5],
    f ' (x) = 2x + 5.
    For critical points, f ' (x) = 0, namely x = - 5⁄2.
    Therefore, at the critical point f (- 5⁄2) = -9 1⁄4.
    At the end points f (-4) = -7 and f (5) = 47.
    Hence, the global minimum value is -9 1⁄4 and the global maximum value is 47.

    (b) curve: f (x) = x⁄x2 + 1     on [-3, 3]
    For critical point, f ' (x) = 0,
    1 - x2 ⁄(x2 + 1)2 = 0,
    1 - x2 = 0.
    x = ±1.
    At the critical points, f (1) = 1⁄2 and f (-1) = - 1⁄2.
    At the end points, f (-3) = - 3⁄10 and f (3) = 3⁄10.
    Hence, the global minimum value is - 1⁄2 and the global maximum value is 1⁄2.
    First Derivative Test for Local Maximum and Local Minimum
    Suppose that c is a critical point of a continuous function f,
    and that differentiable at every point in some interval containing c except possibly at c itself.
    1. If f ' changes from negative to positive at c , then f has a local minimum at c .
    2. If f ' changes from positive to negative at c, then f has a local minimum at c .
    3. If f ' does not change sign at c (f ' is positive on both sides of c or negative on
      both sides of c ), then f has no local maximum or local minimum at c
      4. .
    Example 9.
    Find the critical points of f (x) = x3 - 3x + 1.
    Find the open intervals on which the function is increasing or decreasing. Identify the function's local maximum and minimum.
    Solution
    f (x ) = x3 - 3x + 1.
    f ' (x) = 3x 2 - 3 = 3(x + 1)(x - 1).
    For critical point, f ' (x) = 0, namely, x = ±1.

    Interval x < -1 -1 < x < 1 x > 1
    x + 1 - + +
    x - 1 - - +
    sign of f ' (x) + - +
    Behavior of f   increasing   decreasing   increasing 

    Thus the local maximum at x = -1 and local minimum at x = 1.

    Example 10.
    Show that f(x) = x3 is increasing on both sides of critical point.
    Solution
    curve: f(x) = x3,
    f ' (x) = 3x2
    For critical point, f ' (x ) = 0, namely x = 0.

    Interval x < 1 x > 1
    x2 + +
    sign of f ' (x) + +
    Behavior of f   increasing   increasing  

    Thus the function is increasing on both sides of critical pint.

    Exercise 9.2

    1. Consider the following graphs.

      2. View post on imgur.com
      View post on imgur.com
      (a) State all open intervals where the function is increasing.
      (b) State all open intervals where the function is decreasing.

    2. Find the critical points:

      3. (a) f (x) = x2 - 2x + 5
      (b) f (x) = x3 + 6x2 + 3x + 10
      (c) f (x) = 1 ⁄ x - 3
      (d) f (x) = (x2 - 2x) 2 ⁄ 3
      (e) f (x ) = 1 ⁄ x 2 - 2x + 1
      (f) f (x ) = x 2 ⁄ x 2 + 2

    3. Find the global minimum and global maximum values of the following functions:

      4. (a) f (x) = 3x - x3 + 2 on [-5, 5]
      (b) f (x) = sin 2x on [- π ⁄ 4 , π ⁄ 4 ].

    4. Find the open intervals on which the following functions are increasing or decreasing. Identify the function's local maximum and minimum.

      5. (a) f (x) = x3 - 3x2 + 5
      (b) f (x) =(x2 - 2x) 2 ⁄ 5
      (c) f (x) = x2 + 5 ⁄x
      (d) f (x) = -x3 - 6x2 + 12
      (e) f(x) = x - 3 ⁄ x - 1
      (f) f (x) = x 2 ⁄ 3 (x2 - 16)

    Concavity

    We now introduce the concavity of the graph of functions and then define the point of inflection. Next, we describe the second derivative test to find the local maximum and local minimum.
    The nature of f ' (x) can determine when the graph of a function is bending up or down. More precisely, when f ' (x) is increasing on (a, b), the slope of the tangent line to the curve increases, and the graph bends upward. Similarly, when f ' (x) is decreasing on (a, b), the slope of the tangent line to the curve decreases, and the graph bends downward.

    Definition
    Let f be differentiable on an open interval I. The graph of f is
    1. concave up on I if f ' is increasing on the interval.
    2. concave down on I if f ' is decreasing on the interval.

    View post on imgur.com

    If f '' (x) > 0 for all x in the interval (a, b), then f '(x) is increasing on (a, b). So the graph of f is concave up on (a, b).
    Again, if f '' (x) < 0 for all x in the interval (a, b), then f ' (x) is decreasing on (a, b). So the graph of f is concave down on (a, b).
    Therefore we can use the second derivative to decide whether the function is concave up or concave down.
    Test for Concavity
    Let f be twice differentiable on an open interval I.
    1. If f'' (x) > 0 for all x in I, then f is concave up on I.
    2. If f''(x) < 0 for all x in I, then f is concave down on I.


    View post on imgur.com


    View post on imgur.com

    Example 11.
    Determine the open interval on which the graph of the function f (x) = 1⁄x 2 + 1 is concave up or concave down.
    Solution
    f (x) = 1⁄x 2 + 1 ,
    f ' (x) = - 2x⁄(x 2 + 1)2 .
    We calculate that
    f '' (x) = -2(x2 + 1)2 + 2x(2)(x2 + 1)(2x)⁄ (x2 + 1)4
    = -2(x2 + 1)2 + 8x2(x2 + 1)⁄ (x2 + 1)4
    = 2(3x2 - 1)⁄(x2 + 1)3
    f '' (x) is defined for all real number and f '' (x) = 0.
    Hence,
    2(3x2 - 1)⁄(x2 + 1)3 = 0
    2(3x2 - 1) = 0
    (x + 1⁄ √ 3 ) (x - 1⁄ √ 3 ) = 0.
    Therefore x = ± 1⁄ √ 3
    Interval x < -1⁄ √ 3 -1⁄ √ 3 < x < 1⁄ √ 3 x > 1⁄ √ 3
    x + 1⁄ √ 3 - + +
    x - 1⁄ √ 3 - - +
    sign of f '' (x) + - +
    Behavior of f   concave up   concave down   concave up  

    We now describe a behavior of a point at which the concavity changes.
    Consider the function f (x) = x3 .
    For critical point,
    f ' (x) = 0,
    3x2 = 0
    Thus x = 0.
    Now f '' (x) = 6x. So f '' (x) < 0 for x < 0 and f '' (x) > 0 for x > 0.
    This show that the concavity changes at (0, 0) and the tangent to this curve crosses the curve at (0, 0).
    View post on imgur.com

    Definition
    A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection. At a point of inflection, (c, f (c)), either f '' (c) = 0 or f '' (c) is undefined.
    View post on imgur.com

    Remark. We have an exceptional case that a point c is not inflection point even if f '' (c) = 0. For example,
    f (x) = (x -c)2k
    where the integer k ≥ 2.

    Example 12.
    Determine the open interval on which the graph of following functions
    (a) f (x) = x 1⁄3
    (b) f (x) = 9⁄88 x 11⁄3
    is concave up or concave down. State their inflection points.
    Solution
    (a) f (x) = x 1⁄3
    f ' (x) = 1⁄3 x -2⁄3 ,
    f '' (x) = - 2⁄9 x -5⁄ 3 = - 2⁄9x 5⁄3 .
    Interval x < 0 x = 0 x > 0
    sign of f '' (x) + undefined -
    Behavior of f   concave up   inflection point   concave down  

    Therefore on x < 0 the graph of function is concave up and on x > 0 the graph of function is concave down. The point (0, 0) is an inflection point.

    (b) f (x) = 9⁄88 x 11⁄3
    f ' (x) = 3⁄ 8 x 8⁄3 ,
    f '' (x) = x 5⁄3 .
    Interval x < 0 x = 0 x > 0
    sign of f '' (x) - f '' (0) = 0 +
    Behavior of f   concave down   inflection point   concave up  

    Therefore on x < 0 the graph of function is concave down and on x > 0 the graph of function is concave up. The point (0, 0) is an inflection point.


    The following example shows that concavity of the curve does not change at (0, 0) even if f '' (0) = 0.

    Example 13.
    Show that x = 0 is not inflection point of f (x) = x6.
    Solution
    curve: f (x) = x6,
    f ' (x) = 6x5.
    View post on imgur.com

    For critical points,
    f ' (x) = 0, namely x = 0.
    Next, we consider f '' (x) = 30x4.
    At x = 0, f '' (0) = 0.
    We observed that
    f '' (x) > 0 for x > 0, and f '' (x) > 0 for x < 0,
    so the second derivative does not change sign at x = 0.
    Therefore, x = 0 is not inflection point of f (x) = x6.

    Second Derivative Test

    We now discuss the use of second derivative to determine whether a function is maximum or minimum without looking for sign changes in f ' at critical points.(see Figure in the previous definition).
    Second Derivative Test for Local Maximum and Local Minimum
    Suppose f '' is continuous on an open interval that contains x = c.
    1. If f ' (c) = 0 and f '' (c) > 0, then f has a local minimum at x = c.
    2. If f ' (c) = 0 and f '' (c) < 0, then f has a local maximum at x = c.
    3. If f ' (c) = 0 and f '' (c) = 0, then the test fails. The function f may have
      a local maximum, a local minimum or neither.

    Example 14.
    Find and classify the critical points of f (x) = x4 - 4x3 + 5.
    Solution
    curve: f(x) = x4 - 4x3 + 5
    f ' (x) = 4x3 - 12x2
    = 4x2(x - 3).
    For critical points, f ' (x) = 0, namely x = 0 and x = 3.
    When x = 0, f (0) = 5, and when x = 3, f (3) = -22.
    Therefore, the critical points are (0, 5) and (3, -22).
    Next, f '' (x) = 12 x2 - 24x
    = 12x(x - 2).
    At x = 0, f '' (0) = 0. The second derivative test fails.
    It is observed that f ' (x) < 0 when x < 0 and f ' (x) < 0 when 0 < x < 3.
    Therefore (0, 5) is a point of inflection.
    At x = 3, f '' (3) = 36.
    Therefore (3, -22) is a local minimum.

    Example 15.
    Find the range of f (x) = ln x - x.
    Solution
    curve: f (x) = ln x - x,
    f ' (x) = 1⁄x - 1.
    f '' (x) = - 1⁄x2.
    For critical point, f ' (x) = 0,
    1⁄x - 1 = 0.
    Therefore x = 1
    and
    f (1) = ln 1 - 1 = -1.
    At x = 1, f '' (1) = -1 < 0.
    Thus (1, -1) is a maximum point.
    The range of f (x) is {y : y ≤ -1}.

    Example 16.
    Find the least amount of material needed to build an open cylindrical vessel with a capicity of 400π cm3
    Solution
    Let r = radius and h = height of cylindrical vessel. The volume of vessel is
    πr2h = 400 π
    h = 400⁄r2.
    Since vessel is open top, area of material for vessel is
    A = πr2 + 2πrh + 2πr 400⁄r2
    = πr2 + 800π⁄r, (r > 0).
    Differentiating both sides with respect to r, we get
    dA⁄dr = 2πr - 800π ⁄r2.
    For critical point, dA⁄dr = 0
    2πr - 800π ⁄r2 = 0
    r3 = 400
    r = ∛ 400 .
    Next, d2 A⁄dr2 = 2π + 1600π⁄r3.
    When r = ∛ 400 , d2A⁄dr2 = 2π + 4π > 0.
    Therefore the area of material is the least when r = ∛ 400  .
    Therefore the least amount of material
    = πr2 800π⁄r
    = 1200π⁄ ∛ 400 cm3.

    Exercise 9.3

    1. Determine the open intervals on which the graph of the following functions are concave up or concave down. State their points of inflection.
      2. (a) f (x) = (x - 1)5,

      (b) g (x) = x4 - 2x3

      (c) h (x) = sin x       on [π, -π]

    2. Find and classify the critical points of the following functions.
      3. (a) f (x) = x4 - 4x3 + 5

      (b) g (x) = x5 - 5x + 5

      (c) h (x) = xe-x

    3. Find the range of f (x) = x - ex.


    4. If a piece of string of fixed length is made to enclose a rectangle, show that the enclosed area is the greatest when the rectangle is a square.


    5. Find the minimum value of the sum of a positive number and its reciprocal.


    6. A rectangular field is surrounded by a fence on three of its sides and a straight hedge on the fourth side. If the length of the fence is 320 meters, find the maximum area of the field enclosed.


    7. A rectangular box has a square base of side x cm. If the sum of one side of the square and height is 15 cm, express the volume of the box in terms of x. Use this expression to determine the maximum volume of the box.

    8. Two sides of triangle have a lengths x and y, and the angle between them is θ .
      What volume of θ will maximize the triangle's area?

    9.3 Linearization

    When we have a complicated function that can be impossible to make a perfactly accurate value, we use the linearization that are based on its tangent lines.
    Consider the tangent to the curve y = f (x) that lies closed to the curve near the point of tangency. For a small interval to either side of the point of tangency, the y-values along the tangent line give good approximations to the y-values of the curve.
    For example, we can see this phenomenon by zooming in on the graph of y = x2 + 1⁄2.

    View post on imgur.com

    View post on imgur.com

    Curve y = x2 + 1⁄2 and its tangent y = x + 1⁄4 are very close near (1⁄2, 3⁄4)

    The equation of the tangent to the curve y = f (x) at x = 0, where f is differentiable and passes through the point (a, f (a)) is
    y = f (a) + f ' (a)(x - a).
    This tangent line is the graph of the linear function
    L (x) = f (a) + f ' (a)(x - a).
    View post on imgur.com

    For as long as L(x) is close to the graph of f, the function L (x) gives a good approximation to f (x).

    Definition
    If f is differentiable at x = a, then the approximation function
    L (x) = f (a) + f ' (a)(x - a).
    is the linearization of f at a. The approximation
    f (x) ≈ L (x)
    of f by L is the standard linear approximation of f at a. The point at x = a is the center of the approximation.
    The difference of f (x) and L (x) is the error of the linearization.

    Example 17.
    Find the linearization of f (x) = x2 + 2x at x = 2. Compare the approximate value and true value at x = 2.5, x = 2.05, and x = 2.005.
    Solution
    f (x) = x2 + 2x,
    f ' (x) = 2x + 2
    Since f (2) = 8 and f ' (2) = 6, the linearization is
    L (x) = f (2) + f ' (2)(x - 2)
    = 8 + 6(x - 2) = 6x - 4.
    That is, x2 + 2x ≈ 6x - 4.
    At x = 2.5, the true value gives x2 + 2x
    = (2.5)2 + 2(2.5) = 11.25 and
    the linearization gives 6x - 4 = 6(2.5) - 4 = 11.
    Therefore the approximate value differs from the true value by less than 0.25.
    At x = 0.25, the true value gives x2 + 2x = (2.05)2 + 2(2.05) = 8.3025 and the linearization gives 6x - 4 = 6(2.05) - 4 = 8.3.
    Therefore the approximate value differs from the true value by less than 0.25 × 10-2.
    At x = 2.005, the true value gives
    x2 + 2x = (2.005)2 + 2(2.005) = 8.030025 and
    the linearization gives 6x - 4 = 6(2.005) - 4 = 8.03.
    Therefore, the approximate value differs from the true value by less than 0.25 × 10-4.

    Note:   From the above example, we see that how accurate the approximation is for some value of x near 2. It turns out that, if the variable x moves a small increment, then the true value and approximate value become close.


    Example 18.
    If the radius of a circle increases from r = 5 cm to 5.01 cm, find the approximate increase in the area. Estimate the area of the enlarge circle and find the error.
    Solution
    r=5 δr = 0.01
    Let A be the area of the circle of radius r.
    A(r) = πr2,
    A ' (r) = 2πr.
    Since A (5) = 25π and A ' (5) = 10π, the linearization is
    L (r) = A (5) + A ' (5)(r - 5) = 10πr - 25π.
    That is, πr2 ≈ 10πr - 25π.
    At r = 5.01, L (r) = L (5.01)
    = 10π (5.01) - 25π = 25.1π
    The area of circle of radius 5.01 cm is approximately 25.1π cm2. Hence the approxinate increase is 0.1π cm2.
    The true area is
    A (5.01)= π(5.01)2 = 25.1001π cm2
    Therefore
    Error = True area - Approximate area
    = 0.0001π cm2.

    Example 19.
    Given that f (x) = x1⁄2 , determine the approximate value for √ 101   by using approximation.
    Solution
    f (x) = x1⁄2 .
    When x = 100, f (100) = 1001⁄2 = 10, we have to approximate √ 101  .
    $$ f'(x) = \frac{1}{2\sqrt{x}} $$ $$ f'(100) = \frac{1}{20}. $$ The linearization is
    L (x) = f (100) + f ' (100)(x - 100)
    = x⁄20 + 5.
    That is, √ x   ≈ x⁄20 + 5.
    At x = 101, L (x) = L (101) = 10.05.
    Hence,
    √ 101   ≈ 10.05.

    Exercise 9.4

    1. Find the linearization of

      1. f (x) = √ 1 + x   at x = 3. Compare the approximate value and true value at x = 3.2, x = 3.02 and x = 3.002.

      2. f (x) = x ⁄ x + 1 at x = 1.
        Compare the approximate value and true value at x = 1, x = 1.01, x = 1.001.


    2. Show that the linearization of f (x) = (1 + x)k at x = 0 is L (x) = 1 + kx.

    3. Find the approximate change in the volume of a sphere when its radius decrease from 5 cm to 4.97 cm . Estimate the area of the contract circle and find the error.

    4. If y = 4 √ x   + 3x2, find the approximate change in y when x changes from 9 to 8.98.

    5. If y = 3√ 72 + x2   find the approximate change in y when
      (i) x increases from 3 to 3.01,
      (ii) x decreases from 3 to 2.98.
      Find their errors.

    6. Use linearization to approximate the following values.
      $$ \text{(i)}\ \sqrt{80} $$ $$ \text{ii} \ \sqrt[3]{65} $$ $$ \text(iii) \ \frac{1}{\sqrt{1.22}} $$

  • 9.1 Tangent Line and Normal Line ကိုပြန်သွားရန်

    • Contents

    Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11

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