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The sum of the global minimum and global maximum values of the function $f(x)=\frac{4}{3} x^3-4 x$ in $[0,2]$ is
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The correct answer is:
$0$
Given $f(x)=\frac{4}{3} x^3-4 x, x \in[0,2]$
Now, $\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 4 \mathrm{x}^2-4=0 \Rightarrow \mathrm{x}= \pm 1$
Now $\mathrm{f}(0)=0, \mathrm{f}(1)=\frac{4}{3}-4=-\frac{8}{3}=$ global min
$f(-1)=\frac{-4}{3}+4=\frac{8}{3}$
$f(2)=\frac{4}{3} \times 8-8=\frac{8}{3}$ global max.
So sum of global max + global $\min$
$=\frac{8}{3}-\frac{8}{3}=0$
Now, $\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 4 \mathrm{x}^2-4=0 \Rightarrow \mathrm{x}= \pm 1$
Now $\mathrm{f}(0)=0, \mathrm{f}(1)=\frac{4}{3}-4=-\frac{8}{3}=$ global min
$f(-1)=\frac{-4}{3}+4=\frac{8}{3}$
$f(2)=\frac{4}{3} \times 8-8=\frac{8}{3}$ global max.
So sum of global max + global $\min$
$=\frac{8}{3}-\frac{8}{3}=0$
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