Question

Question #d8374

Answer 1

There is one at `x = 0` and one at `x = 3/2`, and one at `x = -3/2`

Explanation:

The critical or "stationary" points of a function are where the first derivative of that function are zero or undefined. (In our derivative `4x^3-9`, it is impossible to get it to be undefined.)

Therefore, we set our derivative to equal zero.
`4x^3-9x=0`
=>`x(4x^2-9)=0`
We can see that the derivative would equal zero when `x=0` or `4x^2-9=0`

We now solve `4x^2-9=0`
`4x^2-9=0` We use the quadratic formula.
`x=(-0+-sqrt (0^2-4(4)(-9)))/(2(4))`
=>`x=(0+-sqrt 144)/8`
=>`x=(0+-12)/8`
=>`3/2=x=-3/2`

Therefore, the critical points are at when `x=0`, `x=3/2`, and `x=-3/2`