Question

if a function $f(x)$ is continuous on $a,b$ and differentiable on $(a,b)$, then the mean value theorem says that there is at least one number $c$ in the interval $(a,b)$ such that $f(c)=\frac{f(b)-f(a)}{b - a}$. find all possible value(s) for $c$ given $f(x)=x^{3}-8x + 3, -3leq xleq3$. enter your answer(s) separated by commas.
$c =$

Explanation:

Step1: Calculate \(f(a)\) and \(f(b)\)

Here \(a=-3\), \(b = 3\). First, find \(f(-3)\) and \(f(3)\).

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Step2: Calculate \(\frac{f(b)-f(a)}{b - a}\)

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

Differentiate \(f(x)=x^{3}-8x + 3\) with respect to \(x\). Using the power - rule \((x^{n})'=nx^{n - 1}\), we get \(f'(x)=3x^{2}-8\).

Step4: Solve for \(c\)

Set \(f'(c)=1\), so \(3c^{2}-8 = 1\).

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Answer:

\(-\sqrt{3},\sqrt{3}\)