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

Question

Sadie Ackerman · Accepted Answer

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)

Anna Allen · Answer

undefined The function $ f(x) = x^3 $ has local extrema at critical points where the derivative is equal to zero or undefined.

First, let's find the derivative of $ f(x) $:
$$ f'(x) = 3x^2 $$

Now, let's find the critical points by setting the derivative equal to zero:
$$ 3x^2 = 0 $$
$$ x^2 = 0 $$
$$ x = 0 $$

So, the critical point is at $ x = 0 $.

To determine whether $ x = 0 $ is a local maximum, minimum, or neither, we can use the first derivative test.

Considering the sign of the derivative on either side of the critical point:
- When $ x < 0 $, $ f'(x) $ is negative, indicating a decreasing function.
- When $ x > 0 $, $ f'(x) $ is positive, indicating an increasing function.

Therefore, $ f(x) = x^3 $ has a local minimum at $ x = 0 $.

There are no other critical points over the interval $[-10, 10]$, so $ x = 0 $ is the only local extremum of $ f(x) = x^3 $ within this interval.