Finding Local Extrema Points
To find the local maximum and minimum values of the function $f(x) = 2x^3-9ax^2+12a^2x+1$, we first need to find its derivative $f'(x)$.
The derivative is:
$f'(x)$ = $\frac{d}{dx}(2x^3-9ax^2+12a^2x+1)$
$f'(x)$ = $6x^2 - 18ax + 12a^2$
The local extrema occur where $f'(x) = 0$. Setting the derivative to zero:
$6x^2 - 18ax + 12a^2 = 0$
Divide by 6:
$x^2 - 3ax + 2a^2 = 0$
Factor the quadratic equation:
$(x-a)(x-2a) = 0$
The critical points are $x = a$ and $x = 2a$. We need to determine which corresponds to the local maximum (p) and which to the local minimum (q).
Determining Maximum and Minimum Points
We use the second derivative test. First, find the second derivative $f''(x)$:
$f''(x)$ = $\frac{d}{dx}(6x^2 - 18ax + 12a^2)$
$f''(x)$ = $12x - 18a$
Evaluate $f''(x)$ at the critical points:
- At $x = a$: $f''(a) = 12(a) - 18a = -6a$. Since $a > 0$, $f''(a) < 0$. This indicates a local maximum. So, $p = a$.
- At $x = 2a$: $f''(2a) = 12(2a) - 18a = 24a - 18a = 6a$. Since $a > 0$, $f''(2a) > 0$. This indicates a local minimum. So, $q = 2a$.
Calculating the Value of 'a'
We are given the condition $p^2 = q$. Substitute the values of $p$ and $q$ we found:
$a^2 = 2a$
Rearrange the equation:
$a^2 - 2a = 0$
Factor out $a$:
$a(a-2) = 0$
Since we are given that $a > 0$, the only valid solution is $a = 2$.
Calculating the Final Function Value
Now substitute $a = 2$ back into the original function $f(x)$:
$f(x) = 2x^3 - 9(2)x^2 + 12(2^2)x + 1$
$f(x) = 2x^3 - 18x^2 + 12(4)x + 1$
$f(x) = 2x^3 - 18x^2 + 48x + 1$
Finally, calculate the value of $f(3)$:
$f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1$
$f(3) = 2(27) - 18(9) + 144 + 1$
$f(3) = 54 - 162 + 144 + 1$
$f(3) = -108 + 144 + 1$
$f(3) = 36 + 1$
$f(3) = 37$
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