The function f(x)=2 x^3-9 a x^2+12 a^2 x+1 (a | vedclass

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(B) Given the function $f(x)=2 x^3-9 a x^2+12 a^2 x+1$ with $a>0$.First,find the derivative $f^{\prime}(x)$:$f^{\prime}(x)=6 x^2-18 a x+12 a^2$$f^{\pr

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$2$

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(B) Given the function $f(x)=2 x^3-9 a x^2+12 a^2 x+1$ with $a>0$.First,find the derivative $f^{\prime}(x)$:$f^{\prime}(x)=6 x^2-18 a x+12 a^2$$f^{\prime}(x)=6(x^2-3 a x+2 a^2)=6(x-a)(x-2 a)$For local maxima or minima,set $f^{\prime}(x)=0$:$6(x-a)(x-2 a)=0 \Rightarrow x=a, 2 a$Using the second derivative test,$f^{\prime\prime}(x)=12 x-18 a$:At $x=a$,$f^{\prime\prime}(a)=12 a-18 a=-6 a 0$),so $x=a$ is a point of local maxima.At $x=2 a$,$f^{\prime\prime}(2 a)=24 a-18 a=6 a > 0$ (since $a>0$),so $x=2 a$ is a point of local minima.Thus,$p=a$ and $q=2 a$.Given the condition $p^2=q$,we have:$a^2=2 a$$a^2-2 a=0$$a(a-2)=0$Since $a>0$,we get $a=2$.

The function $f(x)=2 x^3-9 a x^2+12 a^2 x+1$ $(a>0)$ attains its maximum and minimum at $p$ and $q$ respectively and $p^2=q$. Then,$a=$

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