Question

f(x)=x^4-1, what is the turning point? Is it 0, -1 or 1?

Answer 1

Well, to find the turning point you differentiate it.
\(f'(x) = 4x^3\)
then set it to 0 \(4x^2=0\) => \(x=0\) then you substiute it into the equation to find \(y\)

Answer 2

Thanks! I wasnt sure if it was to set it completely to 0 or set it to 1, because the x would equal 0 so it would be x=1... but thanks

Answer 3

then forget about the \(y-value\) for the turning point.

Answer 4

don't forget*

Answer 5

Very true. thanks again

Answer 6

np