In the interval $[-2,4]$, the absolute maximum of $f(x)=2 x^3-3 x^2-12 x+5$ occurs $x=$

Question

Aicharya · Accepted Answer

Given, function $f(x)=2 x^3-3 x^2-12 x+5$ So, $f^{\prime}(x)=6 x^2-6 x-12=0$ [for maxima and minima $f^{\prime}(x)=0$ ] $\begin{aligned} & \Rightarrow \quad x^2-x-2=0 \Rightarrow \quad(x+1)(x-2)=0 \ & \Rightarrow \quad x=-1,2 \in[-2,4] \ & \because \quad f(-2)=-16-12+24+5=1 \ & f(-1)=-2-3+12+5=12 \ & f(2)=16-12-24+5=-15 \ & \text {and } \quad f(4)=128-48-48+5=37 \end{aligned}$ The absolute maximum of $f(x)$ occurs at $x=4$. Hence, option (a) is correct.

In the interval \([-2,4]\), the absolute maximum of \(f(x)=2 x^3-3 x^2-12 x+5\) occurs \(x=\)

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