Question

What are the critical values of `f(x)=4x^2-6x`?

Answer 1

The critical values are the roots of the first derivative hence

`f'(x)=0=>8x-6=0=>x=6/8=>x=3/4`

Because `f''(x)=8>0` the value of `x=3/4` is a minimum value

hence `f(3/4)=4*(3/4)^2-6*3/4=>f(3/4)=9/4-18/4=-9/4`