Question
Asked Nov 5, 2019
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f(x)=2x^3-(15x^2)-36x+7 on the interval [-6,9] 

 

By the Mean Value Theorem, we know there exists at least one c in the open interval (−6,9) such that f′(c) is equal to this mean slope. Find all values of c that work and list them.

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Expert Answer

Step 1

Given function

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Step 2

To find the value of c which satisfies the mean value theorem

Step 3
f(x) 2x3-(15x2)-36x +7, [-6,9]
f(x) - 6х* -30х - 36..........)
according tomean value theorem
iff(x)is continuous on close interval[a, b]
and f(x)is differential on the open interval (a, b)
Then there is a number c such that a < c < band
f(b)-f(a)
f '(c)=
b-а
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f(x) 2x3-(15x2)-36x +7, [-6,9] f(x) - 6х* -30х - 36..........) according tomean value theorem iff(x)is continuous on close interval[a, b] and f(x)is differential on the open interval (a, b) Then there is a number c such that a < c < band f(b)-f(a) f '(c)= b-а

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Calculus

Functions

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