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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)=3x525x3+60x on [2,3]f(x)=3 x^{5}-25 x^{3}+60 x \text { on }[-2,3]


Step 1
1 of 2

Differentiating f(x)=3x525x3+60xf(x)=3x^5-25x^3+60x on [2,3][-2,3], we have

f(x)=15x475x2+60\begin{align*} f'(x)&= 15x^4-75x^2+60 \end{align*}

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

f(x)=015x475x2+60=015(x45x2+4)=015(x1)(x+1)(x2)(x+2)=0x=1orx=1orx=2orx=2\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 ff at the critical points and endpoints.

f(2)=3(2)525(2)3+60(2)=16f(1)=3(1)525(1)3+60(1)=38f(1)=3(1)525(1)3+60(1)=38f(2)=3(2)525(2)3+60(2)=16f(3)=3(3)525(3)3+60(3)=234\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)=234f(3)=234, which is the absolute maximum value of ff on [2,3][-2,3]. The least of these values is f(1)=38f(-1)=-38, which is the absolute minimum value of ff on [2,3][-2,3].

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