Question

Over the x-value interval `[-10, 10]`, what are the local extrema of `f(x) = x^3`?

Answer 1

1. Find the derivative of the given function.
2. Set the derivative equal to 0 to find the critical points.
3. Also use the endpoints as critical points .

4a. Evaluate the original function using each critical point as an input value.

OR

4b. Create a sign table/chart using values between the critical points and record their signs .

5.Based on the results from STEP 4a or 4b determine if each of the criticals points are a maximum or a minimum or an inflections points.

Maximum are indicated by a positive value, followed by the critical point, followed by a negative value.

Minimum are indicated by a negative value, followed by the critical point, followed by a positive value.

Inflections are indicated by a negative value, followed by the critical point, followed by negative OR a positive value, followed by the critical point, followed by positive value.

STEP 1:

`f(x)=x^3`

`f'(x)=3x^2`

STEP 2:

`0=3x^2`

`0=x^2`

`sqrt(0)=sqrt(x^2)`

`0=x ->`Critical Point

STEP 3:

`x = 10 ->` Critical Point

`x=-10 ->` Critical Point

STEP 4:

`f(-10)=(-10)^3=-1000`, Point (-10,-1000)

`f(0)=(0)^3=0`, Point (0,0)

`f(10)=(10)^3=1000`, Point (-10,1000)

STEP 5:

Because the result of f(-10) is the smallest at -1000 it is the minimum.
Because the result of f(10) is the largest at 1000 it is the maximum.
f(0) has to be an inflection point.

OR

Check of my work using a signs chart

`(-10)---(-1)---0---(1)---(10)`

`-1` is between critical points `-10` and `0.`

`1` is between critical points `10` and `0.`

`f'(-1)=3(-1)^2=3->positive`

`f'(1)=3(1)^2=3->positive`

The critical point of `0` is surrounded by positive values so it is an inflection point.

`f(-10)=(-10)^3=-1000-> min`, Point (-10,-1000)

`f(0)=(0)^3=0 ->`inflection, Point (0,0)

`f(10)=(10)^3=1000-> max`, Point (-10,1000)