Question

consider the function $f(x) = 3 - 5x^2$ on the interval $-5, 6$. find the average or mean slope of the function on this interval, i.e.\\(\frac{f(6) - f(-5)}{6 - (-5)} = \square\\)\
by the mean value theorem, we know there exists a $c$ in the open interval $(-5, 6)$ such that $f(c)$ is equal to this mean slope. for this problem, there is only one $c$ that works. find it.\
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Explanation:

Step1: Calculate \( f(6) \) and \( f(-5) \)

First, find \( f(6) \): \( f(6) = 3 - 5(6)^2 = 3 - 5\times36 = 3 - 180 = -177 \)
Then, find \( f(-5) \): \( f(-5) = 3 - 5(-5)^2 = 3 - 5\times25 = 3 - 125 = -122 \)

Step2: Compute the mean slope

The mean slope is \( \frac{f(6) - f(-5)}{6 - (-5)} \). Substitute the values: \( \frac{-177 - (-122)}{6 + 5} = \frac{-177 + 122}{11} = \frac{-55}{11} = -5 \)

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

The function \( f(x) = 3 - 5x^2 \), so its derivative \( f'(x) = -10x \)

Step4: Solve \( f'(c) = -5 \)

Set \( -10c = -5 \). Solve for \( c \): \( c = \frac{-5}{-10} = \frac{1}{2} = 0.5 \)

Answer:

The mean slope is \(-5\), and the value of \( c \) is \( \frac{1}{2} \) (or \( 0.5 \)).