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

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(D) Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.Find the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax

Vedclass · Accepted Answer

$37$

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(D) Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.Find the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0$.Factoring gives $6(x - a)(x - 2a) = 0$,so the critical points are $x = a$ and $x = 2a$.The second derivative is $f''(x) = 12x - 18a$.At $x = a$,$f''(a) = 12a - 18a = -6a  0$),so $x = a$ is a local maximum $(p = a)$.At $x = 2a$,$f''(2a) = 24a - 18a = 6a > 0$,so $x = 2a$ is a local minimum $(q = 2a)$.Given $p^2 = q$,we have $a^2 = 2a$. Since $a > 0$,we get $a = 2$.Substituting $a = 2$ into the function: $f(x) = 2x^3 - 9(2)x^2 + 12(2^2)x + 1 = 2x^3 - 18x^2 + 48x + 1$.Now calculate $f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1 = 2(27) - 18(9) + 144 + 1 = 54 - 162 + 144 + 1 = 37$.

If the function $f(x)=2 x^3-9 a x^2+12 a^2 x+1$,where $a > 0$,attains its local maximum and local minimum values at $p$ and $q$ respectively,such that $p^2=q$,then $f(3)$ is equal to:

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