Mean Value Theorem for AP Calculus
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.)
If f(x)=x2 + 4x – 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) = x2 + 4x – 5 is continuous everywhere since it is polynomial.
- The derivative f '(x) = 2x + 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 2x + 4 = 0 x = – 2, i.e., f '(–2) = 0. (See Figure 7.1-3.)
Let f(x) = – 2x + 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 y1 = – 2x + 2. The zeros of y1 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) = x2 – x – 2. Set f '(x) = 0 x2 – 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.)
The points P(1, 1) and Q(3, 27) are on the curve f(x) = x3. 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) = 3x2 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 c2 = . Since only is in the interval (1, 3), c = . (See Figure 7.1-5.)
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 y1=(x – 1)2/3. The graph y1 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:
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