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$Let g(x)=cos(pix^(2)). Below is Amrita's attempt to write a formal justification for the fact that the equation g^(')(x)=0.4 has a solution where -1 < x < 4. Is Amrita's justification complete? If not, why? Amrita's justification: Polynomial and trigonometric functions are differentiable and continuous at all points in their domain, and [-1,4] is within g^(') s domain. Also, {:[g(-1)=-1" and "],[g(4)=1", so "],[(g(4)-g(-1))/(4-(-1))=0.4.]:} So, according to the mean value theorem, g^(')(x)=0.4 must have a solution somewhere between x=-1 and x=4. Choose 1 answer: (A) Yes, Amrita's justification is complete. (B) No, Amrita didn't establish that the average rate of change of g over [-1,4] is equal to 0.4 . (C) No, Amrita didn't establish that g is differentiable.$

Let $g(x)=\cos \left(\pi x^{2}\right)$ . $\newline$ Below is Amrita's attempt to write a formal justification for the fact that the equation $g^{\prime}(x)=0.4$ has a solution where \( -1

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Grade 7

One-step inequalities: word problems

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Q. Let

g(x)=\cos \left(\pi x^{2}\right)

\newline

Below is Amrita's attempt to write a formal justification for the fact that the equation

g^{\prime}(x)=0.4

has a solution where

-1<x<4

\newline

Is Amrita's justification complete?

\newline

If not, why?

\newline

Amrita's justification:

\newline

Polynomial and trigonometric functions are differentiable and continuous at all points in their domain, and

[-1,4]

is within

g^{\prime}

s domain. Also,

\newline

\begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array}

\newline

So, according to the mean value theorem,

g^{\prime}(x)=0.4

must have a solution somewhere between

x=-1

and

x=4

\newline

Choose

1

answer:

\newline

(A) Yes, Amrita's justification is complete.

\newline

(B) No, Amrita didn't establish that the average rate of change of

g

over

[-1,4]

is equal to

0

4

\newline

g

is differentiable.

Mean Value Theorem Application: Amrita's justification relies on the Mean Value Theorem, which states that if a function is continuous on the closed interval $a, b$ and differentiable on the open interval $a, b$ , then there exists some $c$ in $a, b$ such that $f'(c)$ is equal to the average rate of change of the function over $a, b$ .
Verification of Function Continuity and Differentiability: To apply the Mean Value Theorem to $g(x)=\cos(\pi x^2)$ , we need to verify that $g(x)$ is continuous on the closed interval $[-1, 4]$ and differentiable on the open interval $(-1, 4)$ .
Calculation of Average Rate of Change: The function $g(x)$ is a composition of a polynomial function $x^2$ and the cosine function, both of which are continuous and differentiable everywhere. Therefore, $g(x)$ is continuous on $[-1, 4]$ and differentiable on $(-1, 4)$ .
Incorrect Calculation of Average Rate of Change: Amrita calculates the average rate of change of $g(x)$ over the interval $[-1, 4]$ as $(g(4) - g(-1)) / (4 - (-1))$ . However, Amrita incorrectly uses the values of $g(x)$ instead of $g'(x)$ . The average rate of change should be calculated using $g(x)$ , but the values provided, $g(-1) = -1$ and $g(4) = 1$ , are not correct for $g(x) = \cos(\pi x^2)$ .