Question

Find both the maximum value and the minimum value of $3x^4-8x^3+12x^2-48x-25$ on the interval $[0,3]$.

Answer 1

$f^{{}^{\prime }}\left(x\right)=12x^3-24x^2+24x-48$

Putting this equal to zero
$12x^3-24x^2+24x-48=0$
Using hit and trial, we get
$x=2$ as one of the root
So this can be written as $(x-2)(12x^2+24)=0$
Putting $(12x^2+24)=0$
We can see that the equation has two imaginary root
So only real root or point or critical point is $x=2$
Now let's look at second derivative of the given function.

$f^{{}^{\prime \prime }}\left(x\right)=36x^2-48x+24$;

putting 2 in this we get
$f^{{}^{\prime \prime }}\left(2\right)=362^2-48\times 2+24$
$=72$, which is positive.
Hence, $f\left(x\right)$ will have it's minima at $x=2$.

Minimum value of the function will be
$f\left(2\right)=3\times 2^4-8\times 2^3-48\times 2+25$
$=-39$

We also have to look for maxima, so let's look at two extreme of domain so at $x=0$, function will take a value of $25$ and at $x=3$ it will take a value of $16$. Therefor maxima is at $x=0$
Maximum value will be,
$f\left(0\right)=3\times 0^4-8\times 0^3-48\times 0+25$
$=25$

Minimum value of the given function is $-39$

Maximum value of the given function is $25$