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

Question

Accepted Answer

"

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

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

$ f'(x) = 6x^2 - 18ax + 12a^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) = 2x^3 - 18x^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 $.

"

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

Solution

About this question

Drill 25 more like these. Every day. Free.

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

Solution

About this question

More Application of Derivatives questions

Drill 25 more like these. Every day. Free.