Mathematics

Lecture. 6

Chapter . 4

APPLICATIONS OF DERIVATIVES

By Dr. Mohammed Ramidh

OVERVIEW: This chapter studies some of the important applications of derivatives. We learn how derivatives are used to find extreme values of functions, the Mean Value Theorem, L’Hôpital’s Rule, Newton’s Method.

Extreme Values of Functions

EXAMPLE 1: The absolute extreme of the following functions

on their domains can be seen in Figure.

Local (Relative) Extreme Values:

The First Derivative Theorem for Local Extreme Values

EXAMPLE 2: Find the absolute maximum and minimum values of

ƒ(x) = 𝑥2 on [-2 , 1].

Solution:

The function is differentiable over its entire domain, so the only critical point is where ƒ′(x) = 2x = 0, namely We need to check the function’s values at x = 0 and at the endpoints x = -2 and x = 1:

Critical point value: ƒ(0) = 0

Endpoint values: ƒ(1) = 1

ƒ( -2) = 4

The function has an absolute maximum value of 4 at x = -2 and an absolute minimum value of 0 at x = 0.

EXAMPLE 3: Find the absolute extrema values of

g(t) = 8t - 𝑡4 , on the interval [-2,1].

EXAMPLE 4: Find the absolute maximum and minimum values of

ƒ(x) = 𝑥2 3 , on the interval [-2,3].

Solution: The first derivative

EXERCISES 4.1 1. In Exercises1–3,determine from the graph whether the function has

any absolute extreme values on [a, b].

1. 2. 3. 4.

2. In Exercises1–5, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify

the points on the graph where the absolute extremer occur, and include their coordinates.

1. 2. 3.

4. 5.

3. In Exercises1–4,find the function’s absolute maximum and minimum values and critical point.

1. 2. 3.

4.

4. In Exercises1–5, find the derivative at each critical point and determine the local extreme values.

1. 2. 3.

4. 5.

The Mean Value Theorem

To arrive at this theorem we first need Rolle’s Theorem.

Rolle’s Theorem:

The Mean Value Theorem:

Solution: 𝑓 𝑏 −𝑓(𝑎)

𝑏−𝑎 =f′(c)

22−0

2−0 =

4−0

2 = 2

f′(x)=2x ⇾ f′(c) = 2c = 2⇾ c= 1

EXERCISES 4.2 Find the value or values of c that satisfy the equation

in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–4.

Monotonic Functions and The First Derivative Test

”Increasing Functions and Decreasing Functions:

The only functions with positive derivatives are increasing functions; the only functions with negative derivatives are decreasing functions.

”First Derivative Test for Monotonic Functions:

EXAMPLE 5 : Using the First Derivative Test for Monotonic Functions

Find the critical points of ƒ(x) = 𝑥3 − 12𝑥 − 5 and identify the intervals on which ƒ is increasing and decreasing.

Solution: The function ƒ is everywhere continuous and differentiable. The first derivative,

is zero at x = -2 and x = 2.

These critical points subdivide the domain of ƒ

into intervals (‒∞,−2),(-2,2),and(2,∞) and we

determine the sign of f′ by evaluating ƒ at a point

in each subinterval.The behavior of ƒ is determined by then applying Corollary 3 to each subinterval.

First Derivative Test for Local Extrema

FIGURE , A function’s first derivative tells how the graph rises and falls.

EXAMPLE 6: Using the First Derivative Test for Local Extrema,find the critical points of,

Identify the intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values.

summary:

Fig.13-2

1.

EXERCISES 4.3

1. Answer the following questions about the functions whose derivatives are given in Exercises 1–5:

a. What are the critical points of ƒ?

b. On what intervals is ƒ increasing or decreasing?

c. At what points,if any, does ƒ assume local maximum and minimum values?

1. 2. 3.

4. 5.

2. In Exercises 1–4:

1.

2.

3.

4.

Concavity and Curve Sketching

In this section we see how the second derivative gives information about the way the graph of a differentiable function bends or turns.

”Concavity

”The Second Derivative Test for Concavity

”Points of Inflection

Second Derivative Test for Local Extrema

EXAMPLE 8: Using ƒ′ and ƒ″to Graph ƒ Sketch a graph of the function

ƒ(x) = 𝑥4- 4𝑥3 + 10 . using the following steps.

(a) Identify where the extrema of ƒ occur.

(b) Find the intervals on which ƒ is increasing and the intervals on which ƒ is decreasing.

(c) Find where the graph of ƒ is concave up and where it is concave down.

(d) Sketch the general shape of the graph for ƒ.

(e) Plot some specific points, such as local maximum and minimum points, points of inflection,and intercepts. Then sketch the curve.

(e) Plot the curve’s intercepts (if possible) and the points where y′ and y″ are zero. Indicate any local extreme values and inflection points. Use the general shape as a guide to sketch the curve. See figure. FIGURE 4. The graph of ƒ(x) = 𝑥4‒ 4𝑥3+ 10

”Strategy for Graphing y = ƒ(x)

EXAMPLE 7: Using the Graphing Strategy, Sketch the graph of

ƒ(x) = (𝑥+1)2

1+𝑥2

Solution:

EXERCISES 4.4

1. Identify the inflection points and local maxima and minima of the

functions graphed in Exercises 1–3. Identify the intervals on which the functions are concave up and concave down.

1. 2. 3.

2. Use the steps of the graphing procedure on page 272 to graph the

equations in Exercises 1–5. Include the coordinates of any local extreme points and inflection points.

1. 2. 3.

4. 5.

L’Hôpital’s Rule

L’Hôpital’s Rule (First Form)

L’Hôpital’s Rule (Stronger Form)

EXERCISES 4.6

1. In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit.

1. 2. 3.

2. Use l’Hôpital’s Rule to find the limits in Exercises 1–3.

1. 2. 3.

Newton-Raphson method to find the roots of nonlinear algebraic equation

In this section we study a numerical method, called Newton’s method or the Newton–Raphson method, which is a technique to approximate the solution to an equation ƒ(x) = 0. Essentially it uses tangent lines in place of the graph of y = ƒ(x) near the points where ƒ is zero. (A value of x where ƒ is zero is a root of the function ƒ and a solution of the equation ƒ(x) = 0.)

Procedure for Newton’s Method

EXAMPLE 1: Applying Newton’s Method

EXAMPLE 13: Using Newton’s Method

Find the x-coordinate of the point where the curve y = 𝑥3- x crosses the horizontal line y = 1.

In figure we have indicated that the process in Example 2 might have started at the point B0(3, 23) on the curve, with x0 = 3. Point B0 is quite far from the x-axis, but the Tangent at B0 crosses the x-axis at about (2.12, 0), So x1 is still an improvement over x0 . If we use Equation (1) repeatedly as before, with ƒ(x) = 𝑥3 - x - 1 and ƒ′(x) = 3𝑥2 - 1, we confirm the nine-place solution x7 = x6 = 1.3247 17957 in seven steps.

EXAMPLE 14: Use Newton’s method to estimate the one real solution 𝑥3+ 3x + 1 = 0. of Start with x0 = 0 and then find x2 . solution: f(x)= 𝑥3 + 3𝑥 + 1 ⇾ 𝑓′ 𝑥 = 3𝑥2 + 3

When 𝑥0=0, so 𝑥1= 𝑥𝑛 − f(𝑥𝑛)𝑓′ 𝑥𝑛

𝑥1= 0 −1

3

then 𝑥2 = - 13 ‒

−(1

3)3+3 −

1

3 +1

3(−1

3)2+3

= ‒ 29

90 = ‒ 0.32222

EXERCISES 4.7

7. Use Newton’s method to find the two negative zeros of

8.

9.