Absolute max and minimum values

Question

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

$$f(x) = \frac{1}{2}x^4-\frac{2}{3}x^3-2x^2+3$$ on $[-2,3]$.

Min=

I was able to get the max but the answer I got for min was wrong and I can not get the right answer. I do not know what I am doing wrong. I have tried several times.

Answer 1

$$f'(x)=2x^3-2x^2-4x=2x(x^2-x-2)=2x(x+1)(x-2).$$

Thus, $x_{max}=0$, $x_{min_1}=-1$ and $x_{min_2}=2$.

Id est, $$\min_{[-2,3]}f=\min\{f(-1),f(2)\}=f(2)=-\frac{7}{3}$$ and $$\max_{[-2,3]}f=\max\{f(-2),f(3),f(0)\}=f(-2)=\frac{25}{3}.$$ Done!

Answer 2

we get $$f(-2)=\frac{25}{3},f(3)=\frac{15}{2}$$ and $$f'(x)=2x^3-2x^2-4x$$ solving $$f'(x)=0$$ we get $$x_1=0,x_2=2,x_3=-1$$ you must determine which of these solutions give the minimum