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Asked Oct 26, 2019
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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? Give reasons for your answer.
-3sxs-1
- X
f(x)
2x2
- 3x 3
-1 <xs1
Choose the correct answer.
A. No, f(x) is differentiable at every point in (-3,1) but is not continuous at every point in [-3,1].
B. No, f(x) is continuous at every point in [- 3,1] but is not differentiable at every point in (-3,1)
C. Yes, f(x) is continuous at every point in (-3,1) and differentiable at every point in [-3,1]
D. Yes, f(x) is continuous at every point in [-3,1] and differentiable at every point in (-3,1).
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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? Give reasons for your answer. -3sxs-1 - X f(x) 2x2 - 3x 3 -1 <xs1 Choose the correct answer. A. No, f(x) is differentiable at every point in (-3,1) but is not continuous at every point in [-3,1]. B. No, f(x) is continuous at every point in [- 3,1] but is not differentiable at every point in (-3,1) C. Yes, f(x) is continuous at every point in (-3,1) and differentiable at every point in [-3,1] D. Yes, f(x) is continuous at every point in [-3,1] and differentiable at every point in (-3,1).

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Expert Answer

Step 1

Given: -

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-3 s.xs-1 x2-x f(x) 2x2-3x-3 -1xs1

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Step 2

To prove: -

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To verify means value theorem and select the correct option

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Step 3

Calculation...

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-3 <xs-1 -x It is given that f(x)=- 2x2-3x-3 -1xs

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Tagged in

Math

Calculus

Continuity