Find the values of c guaranteed by the Mean Value Theorem (MVT) for f(x)interval [0, 2].In other words, find c € [0, 2] such that f(c)C ==120) [2 – (0)f(x)dx.=1x + 1 over the

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College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter3: Functions

Section3.3: More On Functions; Piecewise-defined Functions

Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...

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Related questions

Question

Find the values of c guaranteed by the Mean Value Theorem (MVT) for f(x)
interval [0, 2].
In other words, find c € [0, 2] such that f(c)
C =
=
1
2
0) [
2 – (0)
f(x)dx.
=
1
x + 1 over the

Expert Solution

Step 1

$G i v e n f (x) = \frac{1}{2} x + 1$

$W e a r e r e q u i r e d t o f i n d V a l u e s o f c g u a r a n t e e d b y t h e M e a n V a l u e T h e o r e m f o r f (x) o v e r t h e i n t e r v a l [0, 2] u \sin g t h e e q u a t i o n f (c) = \frac{1}{2 - 0} \int_{0}^{2} f (x) d x, c \in [0, 2]$

$T h e f o l l o w i n g f o r m u l a s h a s b e e n u s e d t o s o l v e t h i s q u e s t i o n k \int_{a}^{b} [f (x) + g (x)] d x = k \int_{a}^{b} f (x) d x + k \int_{a}^{b} g (x) d x \int_{a}^{b} k \times f (x) d x = k \times \int_{a}^{b} f (x) d x \int_{a}^{b} x^{n} d x = {(\frac{x^{n + 1}}{n + 1})}_{a}^{b} \int_{a}^{b} 1 d x = {(x)}_{a}^{b} {[f (x)]}_{a}^{b} = f (b) - f (a)$