Question

Find the maximum value of $2x^3-24x+107$ in the interval $[1,3]$. Find the maximum value of the same function in $[-3,-1]$.

Answer 1

Let's solve for Domain $[1,3]$
$f^{{}^{\prime }}\left(x\right)=6x^2-24$

Putting this equal to zero we get
$f^{{}^{\prime }}\left(x\right)=6x^2-24=0$
$\therefore x=\pm 2$

Note: $-2$ not in domain
Now let's look at second derivative of the given function.

$f^{{}^{\prime \prime }}\left(x\right)=12x$

At $x=2$, $f^{{}^{\prime \prime }}\left(2\right)=12\times 2=24$
This is positive for $x=2$

So function will take minima at $x=2$
value will be
$f\left(2\right)=2\times 2^3-24\times 2+107$
$=75$

Now let's work for domain $[-3,-1]$

As we have seen above that the function will have it's critical point at $x=-2$
$f^{{}^{\prime \prime }}(-2)=12\times (-2)=-24$
So function will take maxima at $x=-2$
Value will be
$f(-2)=2\times {-2}^3-24\times (-2)+107$
$=135$

We can check at both the extreme of domain to find minima and minima will occur at $x=-3$ and the value is $125$