If the function f(x)=2x^3-9ax^2+12a^2x+1, where a0, attains its local maximum and local m…

Question

If the function $f(x)=2{x}^{3}-9a{x}^{2}+12{a}^{2}x+1$, 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 :

Accepted Answer

To determine the value of $f(3)$ for the function $f(x)=2{x}^{3}-9a{x}^{2}+12{a}^{2}x+1$, where $a>0$, we follow these steps:

First, find the critical points by setting the derivative equal to zero:

${f}^{'}(x)=6{x}^{2}-18ax+12{a}^{2}=0$

Factoring gives:

$6(x-a)(x-2a)=0$

Thus, the critical points are $x=a$ and $x=2a$.

$x=a$ corresponds to a local maximum.

$x=2a$ corresponds to a local minimum.

According to the problem, ${p}^{2}=q$. Substituting $p=a$ and $q=2a$ gives:

${a}^{2}=2a$

Solving for $a$ gives:

$a(a-2)=0$

Since $a>0$, we have $a=2$.

Now, substitute $a=2$ back into the function:

$f(x)=2{x}^{3}-18{x}^{2}+48x+1$

To find $f(3)$:

$f(3)=2(3{)}^{3}-18(3{)}^{2}+48(3)+1$

Calculate each term:

$2(3{)}^{3}=54$

$18(3{)}^{2}=162$

$48(3)=144$

Thus,

$f(3)=54-162+144+1=37$

So, $f(3)=37$.

Solutions · 1