What are the absolute extrema of $f (x) = 2 x^{3} - 15 x^{2} \in [- 2, 10]$ $f(x)=2x^3-15x^2 in[-2,10]$ ?

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What are the absolute extrema of $f (x) = 2 x^{3} - 15 x^{2} \in [- 2, 10]$ $f(x)=2x^3-15x^2 in[-2,10]$ ?

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Answer 1 · 2016-01-06T15:42:02

Minimum: $- 125$ $-125$ , Maximum: $500$ $500$ .

Explanation:

$f (x) = 2 x^{3} - 15 x^{2}$ $f(x)=2x^3-15x^2$ is continuous on the closed interval $[- 2, 10]$ $[-2,10]$ , so it has a minimum and a maximum on the interval. (Extreme Value Theorem)

The extrema must occur at either an endpoint of the interval or at a critical number for $f$ $f$ in the interval.

Find the critical numbers for $f$ $f$ .

$f'(x) = 6x^2-30x = 6x(x-5)$ .

$f'$ is never undefined and $f'(x)=0$ at $x=0$ and $5$ .

Both $0$ and $5$ are in the interval of interest (the domain of this problem).

Do the arithmetic to find

$f(-2) = -16-25(4) = -16-60 = -76$

$f(0) = 0$

$f(5) = 2(5)^3-15(5)^2 = 10(25)-15(25) = -5(25)=-125$

$f(10) = 2000-1500 = 500$

Minimum: $-125$ , Maximum: $500$ .

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What are the absolute extrema of $f (x) = 2 x^{3} - 15 x^{2} \in [- 2, 10]$ $f(x)=2x^3-15x^2 in[-2,10]$ ?

1 Answer

Explanation:

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What are the absolute extrema of f(x)=2x^3-15x^2 in[-2,10]f(x)=2x3−15x2∈[−2,10]f(x)=2x^3-15x^2 in[-2,10]?

1 Answer

Explanation:

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What are the absolute extrema of $f (x) = 2 x^{3} - 15 x^{2} \in [- 2, 10]$ $f(x)=2x^3-15x^2 in[-2,10]$ ?