Question

calc

Answer 1

x= -1 and 1 right?

Answer 2

To find critical values just find \(f'(x) = 0\)

Answer 3

thats what i did, am i wrong?

Answer 4

\[f(x) = x^4 -6x^2 -5\]\[f'(x) = 4x^3 -12x=0\]\[f'(x): 4x(x^2 -3)=0\]\[f'(x): 4x =0\]\[f'(x) : x^2-3=0\]\[f'(x):x=0\]\[f'(x): x^2=3 \implies x= \pm \sqrt{3}\] there you go.

Answer 5

wow thank you!

Answer 6

\[\boxed {x=0 \ , \ \pm \sqrt{3}}\]

Answer 7

Np :)