Question

consider the function ( f(x)=3 - 2x^{2} ) on the interval ( -3,3 ). find the average or mean slope of the function on this interval, i.e. ( \frac{f(3)-f(-3)}{3-(-3)} ).
by the mean value theorem, we know there exists a ( c ) in the open interval ( (-3,3) ) such that ( f^{prime}(c) ) is equal to this mean slope. for this problem, there is only one ( c ) that works. find it.

Explanation:

Step1: Calculate \( f(3) \) and \( f(-3) \)

Given \( f(x)=3 - 2x^{2} \), then \( f(3)=3-2\times3^{2}=3 - 18=-15 \), \( f(-3)=3-2\times(-3)^{2}=3 - 18=-15 \)

Step2: Calculate the mean - slope

The formula for the mean - slope is \( \frac{f(3)-f(-3)}{3-(-3)} \). Substitute \( f(3)=-15 \) and \( f(-3)=-15 \) into the formula:

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Step3: Find the derivative of \( f(x) \)

Differentiate \( f(x)=3-2x^{2} \) using the power rule \( (x^{n})^\prime=nx^{n - 1} \). So \( f^\prime(x)=-4x \)

Step4: Solve \( f^\prime(c)=0 \)

Set \( f^\prime(c)=-4c = 0 \). Solving for \( c \), we get \( c = 0 \)

Answer:

The mean - slope is \(0\). The value of \(c\) is \(0\)