if a function $f(x)$ is continuous on $a,b$ and differentiable on $(a,b)$, then the mean value theorem says…

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.\n$c =$

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.\n$c =$

Answer

Explanation:

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

Here (a=-3), (b = 3). First, find (f(-3)) and (f(3)). [ \begin{align*} f(-3)&=(-3)^{3}-8\times(-3)+3\ &=-27 + 24+3\ &=0 \end{align*} ] [ \begin{align*} f(3)&=3^{3}-8\times3 + 3\ &=27-24 + 3\ &=6 \end{align*} ]

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

[ \begin{align*} \frac{f(3)-f(-3)}{3-(-3)}&=\frac{6 - 0}{6}\ &=1 \end{align*} ]

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). [ \begin{align*} 3c^{2}&=9\ c^{2}&=3\ c&=\pm\sqrt{3} \end{align*} ]

Answer:

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