Find the interval where the function f(x)=x^(2/3) is increasing and the interval where it is decreasing.

Question

Also, find the maximum or minimum value of f(x). Sketch the graph and label the critical point/s.

Answer 1

We will set the derivative of the function equal to zero, since the slope of the tangent line whose slope is zero lies at the value of the critical points. This is known as the first derivative test.

 

d/dt[x^2/3] = 0

 

(2/3)x^1/3 = 0

 

x = 0

 

The critical point is x=0.

 

To find where the graph increases and decreases, we perform a test point using critical point.  We will use x=-1 and x=1 as our test points.  We substitute the values into the derivative to see if we get a negative or positive slope near the critical point. 

 

d/dt[f(-1)]  = (2/3)(-1)^1/3

                 = (2/3)(-1)

                 = -2/3

 

d/dt[f(1)]  = (2/3)(1)^1/3

                = (2/3)(1)

                = 2/3

 

Based on this test, the graph decreases from the left of the critical point and increases from the right of the critical point.  Therefore, we will have a minimum point.

 

Now we substitute x=0 into the original function to find the value of the minimum point.

 

f(0) = 0^2/3

      = 0

 

The minimum value of the graph is (0,0).