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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.6 - Problem 22e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.6 - Problem 22e

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# Mean Value Theorem a. Determine whether the Mean Value ISBN: 9780321570567 2

## Solution for problem 22E Chapter 4.6

Calculus: Early Transcendentals | 1st Edition

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Problem 22E

Mean Value Theorem a. Determine whether the Mean Value Theorem applies to the following functions on the given interval [a. b]. b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem. c. Make a sketch of the function and the line that passes through (a.f(a)) and (b,f(b)). Mark the points P (if they exist) at which the slope of the function equals the slope of the secant line. Then sketch the tangent line at P. f(x) =x+2 on [-1,2]

Step-by-Step Solution:

Solution 22E Step 1 In this problem we have to check whether the mean value theorem is applicable for x f(x) = x+2on [-1,2]. And if it is applicable we have to find a point that guarantees the mean value theorem and then we have to draw a sketch of the function with given conditions. Mean Value theorem: If f is defined and continuous on the closed interval [a,b]and differentiable on the open interval (a,b)then there is at least one point cin (a,b)that is a < c < bsuch that f (c) = f(b) f(.) ba a. Determine whether the Mean Value Theorem applies to the following functions on the given interval [a,b]. x Given f(x) = x+2 (x+2)1 x(1) f (x) = (x+2)2 = x+2x (x+2) 2 = (x+2)2 Since f (x)is finite for all values of x in the interval (-1,2), f(x)is differentiable in (-1,2) … (1) We know that “Every differentiable function is continuous”. f(x)is continuous in [-1,2] …. (2) From (1) and (2) we get f(x)satisfies all the requirements of mean value theorem. x Hence mean value theorem is applicable for f(x) = x+2 on [-1,2]. Step 3 b. If so, find or approximate the point(s)that are guaranteed to exist by the Mean Value Theorem. x From step 2, we get that mean value theorem is applicable for f(x) = x+2on [-1,2]. Therefore by mean value theorem, There is at least one point cin (-1,2) such that f (c) = f(b) f(… (3) ba x Given f(x) = x+2 (x+2)1 x(1) f (x) = (x+2) = x+2x (x+2)2 = 2 2 (x+2) Substituting x = cwe get f(c) = 2 (c+2) Here a =1and b = 2 Therefore f(a) = f(1) = 1 = 1 1+2 And f(b) = f(2) = 2 = = 1 2+2 4 2 Substituting all value in (3) we get, 1 2 = 2+1 (c+2) 2+1 2 2 (c+2)= 3 2 3 (c+2)= 6 2 1 (c+2)= 2 2 4 = (c+2) We know that (a+b) = a +2ab+b 2 4 = c +4c+4 2 c +4c = 0 c(c+4) = 0 c = 0, c = 4 Since c = 0 (1,2)and c = 4 / (1,2) x We can say that the point that guarantees the mean value theorem for f(x) = x+2 on (-1,2) is 0. Step 4 c. Make a sketch of the function and the line that passes through (a,f(a)) and (b,f(b)). Mark the points P (if they exist) at which the slope of the function equals the slope of the x secant line. Then sketch the tangent line at P. f(x) = x+2 on [-1,2] x The given function is f(x) = x+2 . The point (a,f(a)) = (1, 1)and (b,f(b)) = (2, ) 2 The required graph is given below.

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