Vtyjkyuk

I got the second derivative $12x-6$ for this equation $y = 2 x ^ 3 - 3 x ^ 2 - 36 x + 4 = 0$ and the local minimum is something $3$ and $-2$ as local maximum, but I don't know how to find absolute maximum and minimum.

Am I supposed to use those interval points as $f(0)$ and $f(4)$ and apply to derivative and answer should be $-6$ as absolute maximum and $42$ and absolute minimum. I don't know. Please guide me.

Guide:

Your derivative is not correct. You have computed the second derivative instead.

$$f'(x) = 6x^2-6x-36=0$$

To find relative maxima and minima, set $f'(x) =0$ and solve for $x$. Those are candidate for relative maxima and minima. You can verify which one is it by evaluating the second deritive.

To find absolute maxima and minima in $[0,4]$, compare the values of $f(0), f(4)$, as well as any values for relative maxima and minima that you have obtained earlier if they reside in the interior.

You gave the second derivative of $f(x)=2x^3-3x^2-36x+4$. To find the maximum and minimum we have to find such $x$ with $f'(x)=0$ and then we can use $f''$ to decide if we have a maximum or minimum.

It is $f'(x)=6x^2-6x$ and $6x^2-6x-36=0Leftrightarrow x^2-x-6=0$.

Using the quadratic formula we get:

$x_1,2=frac12pmsqrtfrac14+6=frac12pmsqrt6.25=frac12pm 2.5$

$x_1= 3$ and $x_2=-2$.

Now we have to plug these solutions into $f''$ and see if it is less or greater than zero.
If $f''(x)>0$ we have a minimum and $f''(x)<0$ we have a maximum.