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# Determine the location and value of the absolute extreme values of f on the given interval, if they exist. $f(x)=3 x^{5}-25 x^{3}+60 x \text { on }[-2,3]$

Solution

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Differentiating $f(x)=3x^5-25x^3+60x$ on $[-2,3]$, we have

\begin{align*} f'(x)&= 15x^4-75x^2+60 \end{align*}

To find the critical points, we solve the equation $f'(x)=0$ and we search for values where $f'(x)$ is undefined.

$\begin{gather*} f'(x)=0\\ 15x^4-75x^2+60=0\\ 15(x^4-5x^2+4)=0\\ 15(x-1)(x+1)(x-2)(x+2)=0\\ x=1 \quad or \quad x=-1\quad or \quad x=2 \quad or \quad x=-2 \end{gather*}$

These four critical points and the endpoints are candidates for the location of the absolute extrema.

The next step is to evaluate $f$ at the critical points and endpoints.

\begin{align*} f(-2)&=3(-2)^5-25(-2)^3+60(-2)= -16 \\ f(-1)&=3(-1)^5-25(-1)^3+60(-1)= -38\\ f(1)&=3(1)^5-25(1)^3+60(1)= 38\\ f(2)&=3(2)^5-25(2)^3+60(2)= 16\\ f(3)&=3(3)^5-25(3)^3+60(3)= 234 \end{align*}

The largest of these function values is $f(3)=234$, which is the absolute maximum value of $f$ on $[-2,3]$. The least of these values is $f(-1)=-38$, which is the absolute minimum value of $f$ on $[-2,3]$.

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