Answer :
[tex]f(x)=x^4-32x+4\\
f'(x)=4x^3-32\\\\
4x^3-32=0\\
4(x^3-8)=0\\
x=2\\\\
f(2)=2^4-32\cdot2+4=16-64+4=-44[/tex]
The only extremum is [tex]-44[/tex] and it's minimum.
The only extremum is [tex]-44[/tex] and it's minimum.
