1. Find the slope of the secant line \[\] \[\mbox{Find the value of the function}\] \[\mbox{at the interval endpoints}\] \[f\left(3\right)=9-18+9=0\] \[f\left(8\right)=64-48+9=25\] \[\mbox{Now find the slope of the secant line}\] \[\mbox{that connects those points}\] \[m_{s}=\frac{25-0}{8-3}=5\]
2. Find the derivative \[\] \[f'\left(x\right)=2x-6\]
3. Find \(c\) \[\] Since the function is a polynomial, it is guaranteed to be differentiable over the given interval; therefore, by the MVT there exists at least one value \(c\) such that \(f'(c)=5\). \[\mbox{To find c, set \(f'(c)\) equal to 5}\] \[f'\left(c\right)=2c-6\] \[2c-6=5\] \[2c=11\] \[c=5.5\]

Therefore, at \(x=5.5\) the derivative of the function is equal to the slope of the secant line that connects the endpoints of the given interval. This is proven by the Mean Value Theorem because the original function is differentiable and continuous over the specified interval.

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If allowed to do so, we can check this with a graph.

Graph both the secant line and the tangent line.

Tangent Line: \(y-6.25=5\left(x-5.5\right)\) (blue)

Secant Line: \(y=5x-3\) (green)

*We used our calculator to find value of the function at \(x=5.5\).

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As we see the tangent line and the secant line are parallel.

First find the equation of the secant line that connects the endpoints of the interval.

\[m_{s}=\frac{4-0}{9-1}=\frac{1}{2}\]

\[\therefore\]

We cannot justify using the MVT that there exists at least one value \(c\) on the specified interval such that \(f'(c)=-1\) since the slope of the secant line is \(\frac{1}{2}\); however, we can justify using the MVT that there exists one value \(c\) on the specified interval such that \(f'(c)=\frac{1}{2}\) because the\(f\) is both continuous and differentiable.

1. Find the slope of the secant line \[\] \[\mbox{Find the value of the function}\] \[\mbox{at the interval endpoints}\] \[f\left(-2\right)=3\left(-8\right)-5\left(-2\right)=-14\] \[f\left(2\right)=3\left(8\right)-5\left(2\right)=14\] \[\mbox{Now find the slope of the secant line}\] \[\mbox{that connects those points}\] \[m_{s}=\frac{14-\left(-14\right)}{2+2}=7\]
2. Find the derivative \[\] \[f'\left(x\right)=9x^{2}-5\]
3. Find \(c\) \[\] Since the function is a polynomial, it is guaranteed to be differentiable over the given interval; therefore, by the MVT there exists at least one value \(c\) such that \(f'(c)=7\). \[\mbox{To find c, set \(f'(c)\) equal to 7}\] \[f'\left(c\right)=9c^{2}-5\] \[9c^{2}-5=7\] \[9c^{2}=12\] \[c^{2}=\frac{12}{9}=\frac{4}{3}\] \[c=\pm\sqrt{\frac{4}{3}}=\pm\frac{2}{\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}\]

Therefore, at \(x=\pm\frac{2\sqrt{3}}{3}\) the derivative of the function is equal to the slope of the secant line that connects the endpoints of the given interval. This is proven by the Mean Value Theorem because the original function is differentiable and continuous over the specified interval.