Practice problems for these concepts can be found at: Graphs of Functions and Derivatives Practice Problems for AP Calculus

If *f* is a function that satisfies the following conditions:

*f*is continuous on a closed interval [*a*,*b*]*f*is differentiable on the open interval (*a*,*b*)

then there exists a number *c* in (*a*, *b*) such that *f* '(*c*) = (See Figure 7.1-2.)

**Example 1**

If *f*(*x*)=*x*^{2} + 4*x* – 5, show that the hypotheses of Rolle's Theorem are satisfied on the interval [–4, 0] and find all values of *c* that satisfy the conclusion of the theorem. Check the three conditions in the hypothesis of Rolle's Theorem:

*f*(*x*) =*x*^{2}+ 4*x*– 5 is continuous everywhere since it is polynomial.- The derivative
*f*'(*x*) = 2*x*+ 4 is defined for all numbers and thus is differentiable on (–4, 0). *f*(0) =*f*(–4) = –5. Therefore, there exists a*c*in (–4, 0) such that*f*'(*c*) = 0. To find*c*, set*f*'(*x*) = 0. Thus 2*x*+ 4 = 0*x*= – 2, i.e.,*f*'(–2) = 0. (See Figure 7.1-3.)

**Example 2**

Let *f*(*x*) = – 2*x* + 2. Using Rolle's Theorem, show that there exists a number *c* in the domain of *f* such that *f* '(*c*) = 0. Find all values of *c*.

Note *f*(*x*) is a polynomial and thus *f*(*x*) is continuous and differentiable everywhere.

Enter *y*1 = – 2*x* + 2. The zeros of *y*1 are approximately –2.3, 0.9, and 2.9 i.e., *f*(–2.3) = *f*(0.9) = *f*(2.9) = 0. Therefore, there exists at least one *c* in the interval (–2.3, 0.9) and at least one *c* in the interval (0.9, 2.9) such that *f* '(c) = 0. Use *d* [*differentiate*] to find *f* '(*x*): *f* '(*x*) = *x*^{2} – *x* – 2. Set *f* '(*x*) = 0 *x*^{2} – *x* – 2 = 0 or (*x* – 2)(*x* + 1) = 0.

Thus *x* = 2 or *x* = – 1, which implies *f* '(2) = 0 and *f* '(–1) = 0. Therefore the values ofx = 2 or x = – 1, which implies f '(2) = 0 and f '(–1) = 0. Therefore the values of *c* are –1 and 2. (See Figure 7.1–4.)

**Example 3**

The points *P*(1, 1) and *Q*(3, 27) are on the curve *f*(*x*) = *x*3. Using the Mean Value Theorem, find *c* in the interval (1, 3) such that *f* '(*c*) is equal to the slope of the secant

The slope of secant is *m* = = 13. Since *f*(*x*) is defined for all real numbers, *f*(*x*) is continuous on [1, 3]. Also *f* '(*x*) = 3*x*^{2} is defined for all real numbers. Thus, *f*(*x*) is differentiable on (1, 3). Therefore, there exists a number *c* in (1, 3) such that *f* '(*c*) = 13. Set *f* '(*c*) = 13 3(*c*)^{2} =13 or *c*^{2} = . Since only is in the interval (1, 3), *c* = . (See Figure 7.1-5.)

**Example 4**

Let *f* be the function *f*(*x*) = (*x* – 1)^{2/3}. Determine if the hypotheses of the Mean Value Theorem are satisfied on the interval [0, 2] and if so, find all values of *c* that satisfy the conclusion of the theorem.

Enter *y*1=(*x* – 1)^{2/3}. The graph *y*1 shows that there is a cusp at *x* =1. Thus, *f*(*x*) is not differentiable on (0, 2) which implies there may or may not exist a *c* in (0, 2) such that *f* '(*c*) = . The derivative *f* '(*x*) = Set Note that *f* is not differentiable (*a* + *x* =1). Therefore, *c* does not exist. (See Figure 7.1–6.)

Practice problems for these concepts can be found at:

Graphs of Functions and Derivatives Practice Problems for AP Calculus

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