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AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 5

Lesson 1: Using the mean value theorem

Justification with the mean value theorem

AP.CALC: FUN‑1 (EU), FUN‑1.B (LO), FUN‑1.B.1 (EK)

Let f, left parenthesis, x, right parenthesis, equals, 2, start superscript, x, end superscript, minus, start text, s, i, n, end text, left parenthesis, pi, x, right parenthesis.

Below is Rafael's attempt to write a formal justification for the fact that the equation f, prime, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 4, end fraction has a solution where minus, 2, is less than, x, is less than, minus, 1.

Is Rafael's justification complete? If not, why?

Rafael's justification:
Exponential and trigonometric functions are differentiable and continuous at all points in their domain, and minus, 2, is less than or equal to, x, is less than or equal to, minus, 1 is within f's domain.
So, according to the mean value theorem, f, prime, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 4, end fraction must have a solution somewhere in the interval minus, 2, is less than, x, is less than, minus, 1.

(Choice A)
Yes, Rafael's justification is complete.
(Choice B)
No, Rafael didn't establish that the average rate of change of f over open bracket, minus, 2, comma, minus, 1, close bracket is equal to start fraction, 1, divided by, 4, end fraction.
(Choice C)
No, Rafael didn't establish that f is differentiable.

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AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 5

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