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The maximum value of $2 x^3-24 x+107$ in the interval $[-3,3]$ is
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Verified Answer
The correct answer is:
$139$
Let $f(x)=2 x^3-24 x+107$
At $x=-3, f(-3)=2(-3)^3-24(-3)+107=125$
At $x=3, f(3)=2(3)^3-24(3)+107=89$
For maxima or minima, $f^{\prime}(x)=6 x^2-24=0$ $\Rightarrow x=2,-2$
So at $x=2, f(2)=2(2)^3-24(2)+107=75$
at $x=-2, f(-2)=2(-2)^3-24(-2)+107=139$
Thus the maximum value of the given function in $[-3,3]$ is 139.
At $x=-3, f(-3)=2(-3)^3-24(-3)+107=125$
At $x=3, f(3)=2(3)^3-24(3)+107=89$
For maxima or minima, $f^{\prime}(x)=6 x^2-24=0$ $\Rightarrow x=2,-2$
So at $x=2, f(2)=2(2)^3-24(2)+107=75$
at $x=-2, f(-2)=2(-2)^3-24(-2)+107=139$
Thus the maximum value of the given function in $[-3,3]$ is 139.
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