# Find the local maximum and minimum values of using both the First and Second Derivative Tests.

Find the local maximum and minimum values of  using both the First and Second Derivative Tests. Which  method do you prefer?
f (x) = 1 + 3x^2 - 2x^3.
asked Jan 22, 2015 in CALCULUS
retagged Jan 22, 2015

Step 1:

The function

and

and

The critical points are and the test intervals are .

Step 2:

First derivative test:

 Interval Test Value Sign of Conclusion Decreasing Increasing Decreasing

is changes its sign from negative to positive, hence f  has a local minimum at .

Local minimum is

is changes its sign from positive to negative, hence f  has a local maximum at

Local maximum is .

Solution:

Local minimum is  .

Local maximum is .

Step 1:

The function

Second derivative test:

Differentiate with respect to :

, curve is concave up, thus is a local minimum.

Local minimum is .

, curve is concave down, thus is a local minimum.

Local maximum is .

Solution:

Local minimum is  .

Local maximum is .