Step-by-step explanation:
Derivative of the first function:
[tex]f(x)=2x^{4} -36x^{2} +2 on D:[-4,4][/tex]
[tex]f'(x)=8x^{3} -72x[/tex]
set that equal to 0 and solve for possible x values.
[tex]8x(x^{2} -9)=0\\x=0, -3, 3[/tex]
put x back in to original equation to get a y value
y = 2, -160, -160,
Absolute min at (-3, -160) and (3, 160) and an absolute max at (0,2)
Second Part
Find derivative of [tex]f(x)=2x^{3} +20x^{2} +50x +1[/tex]
[tex]f'(x)=6x^{2} +40x+50[/tex]
set equal to zero, it is a quadratic so you can use the quadratic formula to solve for x
x= -5 or -5/3
put x back into the original function to get a y value
y = 1 or -973/27
so absolute max at (-5, 1) and absolute min at (-5/3, -973/27)
