Question

How do you find the local max and min for `f(x)=x^3-2x+5` on the interval (-2,2)?

Answer 1

See below.

Explanation:

First find the 1st derivative of `x^3-2x+5`

`d/dx(x^3-2x+5)=3x^2-2`

Maximum and minimum points have a horizontal gradient .i.e. gradient of `0`. Solving the 1st derivative will allow us ti identify these points.

`f'(x)=0`

`:.`

`3x^2-2=0=>x=sqrt(6)/3 , x=-sqrt(6)/3`

Plugging these values into the second derivative, will allow us to find whether these are maximum, minimum or points of inflection. Using the following, if:

`f''(x)>0color(white)(8888)` minimum value

`f''(x)<0color(white)(8888)` maximum value

`f''(x)=0color(white)(8888)` maximum/minimum or point of inflection.

The second derivative is the derivative of the first derivative.

`:.`

`d/dx(3x^2-2)=6x`

Plugging in our values of `x`:

`6(sqrt(6)/3)=2sqrt(6)color(white)(8888888)` ( minimum value )

`6(-sqrt(6)/3)=-2sqrt(6)color(white)(88)` ( maximum value )

Both `color(white)(8888)x=sqrt(6)/3 , x=-sqrt(6)/3` are in `( -2 ,2)`

GRAPH:

graph{y=x^3-2x+5 [-5, 5, -20, 20]}