Answer:
Minimum 8 at x=0, Maximum value: 24 at x=4
Step-by-step explanation:
Retrieving data from the original question:
[tex]f(x)=x^{2}+8\:over\:[-1,4][/tex]
1) Calculating the first derivative
[tex]f'(x)=2x[/tex]
2) Now, let's work to find the critical points
Set this
[tex]2x=0\\x=0[/tex]
0, belongs to the interval. Plug it in the original function
[tex]f(0)=(0)^2+8\\f(0)=8[/tex]
3) Making a table x, f(x) then compare
x| f(x)
-1 | f(-1)=9
0 | f(0)=8 Minimum
4 | f(4)=24 Maximum
4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.
