Question

What are the critical points of `f(x) = x^3 − 12x + 7`?

Answer 1

The critical points are:

MIN: `(2;-9)`
MAX: `(-2;23)`

Explanation:

The critical points are the x values where:

`f'(x)=0`

We obtain:

`f'(x)=3x^(3-1)-12x^(1-1)+0=3x^2-12=3*(x^2-4)=`
`=3*(x+2)(x-2)`

Then:

`f'(x)=0 iff 3*(x+2)(x-2)=0`

a multipication is zero if the factors are zero:

`x+2=0`

`x_1=-2`

`x-2=0`

`x_2=+2`

Now, to find if they are min or max you have to evaluate `f'(x)>0`

`f'(x)>0` for `x in ]-oo;-2[ uu ]2;+oo[`

then

`f(x)` is growing for `x in ]-oo;-2[ uu ]2;+oo[`

`f'(x)<0` for `x in ]-2;2[`

then

`f(x)` is decreasing for `x in ]-2;2[`

then
for `x=-2` we have a MAX
for `x=2` we have a MIN

`f(-2)=-8+24+7=23`

`f(2)=8-24+7=-9`