If the function f(x)=2x^3-9ax^2+12a^2x+1, a0 | vedclass

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(A) Given $f(x)=2x^3-9ax^2+12a^2x+1$.Find the derivative: $f'(x)=6x^2-18ax+12a^2$.For local extrema,set $f'(x)=0$,so $6(x^2-3ax+2a^2)=0$,which factors

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$x^2-6x+8=0$

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(A) Given $f(x)=2x^3-9ax^2+12a^2x+1$.Find the derivative: $f'(x)=6x^2-18ax+12a^2$.For local extrema,set $f'(x)=0$,so $6(x^2-3ax+2a^2)=0$,which factors to $6(x-a)(x-2a)=0$.The roots are $x=a$ and $x=2a$.Given the local maximum at $x=\alpha$ and local minimum at $x=\alpha^2$,we have two cases:Case $1$: $\alpha=a$ and $\alpha^2=2a$. Then $a^2=2a \Rightarrow a(a-2)=0$. Since $a>0$,$a=2$.If $a=2$,the roots are $\alpha=2$ and $\alpha^2=4$. The equation is $(x-2)(x-4)=x^2-6x+8=0$.Case $2$: $\alpha=2a$ and $\alpha^2=a$. Then $(2a)^2=a \Rightarrow 4a^2-a=0 \Rightarrow a(4a-1)=0$. Since $a>0$,$a=1/4$.If $a=1/4$,the roots are $\alpha=1/2$ and $\alpha^2=1/4$. The equation is $(x-1/2)(x-1/4)=x^2-(3/4)x+1/8=0$,or $8x^2-6x+1=0$.Checking the second derivative $f''(x)=12x-18a$:For $a=2$,$f''(2)=24-36=-12  0$ (min). This matches.For $a=1/4$,$f''(1/2)=6-18(1/4)=1.5 > 0$ (min) and $f''(1/4)=3-18(1/4)=-1.5 Thus,the correct equation is $x^2-6x+8=0$.

If the function $f(x)=2x^3-9ax^2+12a^2x+1, a>0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$,then $\alpha$ and $\alpha^2$ are the roots of the equation:

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