Example 1: 2nd Derivative Test

Determine all relative extrema of the function using the 2 nd Derivative Test.

f ( x ) = 3 x 5 20 x 3

Step 1: Find the first derivative, f ( x ) , and the second derivative, f ( x ) .

f ( x ) = 3 x 5 20 x 3

f ( x ) = 15 x 4 60 x 2

f ( x ) = 60 x 3 120 x

Step 2: Find the critical values of the equation.

f ( x ) = 0 OR f ( x ) = Does Not Exist ( DNE )

In this situation you do not need to concern yourself with the f ( x ) = Does Not Exist ( DNE ) scenario because there are no fractions, and no other places that math would “break”.

All that is needed is to take the derivative and set it equal to zero, then solve that equation for x . Most times when you set an equation equal to zero, the algebraic method for solving it is the same. Factor the equation and then set those smaller equations equal to zero, and find their solutions.

Algebra Moves:

i) Factor a 15 x 2 out of the equation.

ii) Separate it into two math problems with both pieces being set equal to zero.

iii) First, in the left column we will isolate the x 2 . Start by dividing both sides by 15 ,and then take the square root of both sides to solve for x .

iv) Next, in the right column we will again isolate the x 2 , and then take the square root of both sides to solve for x . Remember when you apply the square root , you get the ± of that result, not just the positive.

0 = 15 x 4 60 x 2

0 = 15 x 2 ) x 2 4 )

0 = 15 x 2

0 15 = 15 x 2 15

0 = x 2

0 = x 2

0 = x

0 = x Critical Value

0 = ( x 2 4 )

0 + 4 ̲ = x 2 4 + 4 ̲

4 = x 2

4 = x 2

± 2 = x

2 = x and 2 = x

2 = x Critical Value and 2 = x Critical Value

Step 3: Plug your critical values into your second derivative, f ( x ) .

You have three critical values in this situation.

x = 0 , x = 2 , and x = 2 .

You will need to plug each of those x-values into our second derivative, f ( x ) .

f ( x ) = 60 x 3 120 x

f ( 0 ) = 60 x ( 0 ) 3 120 ( 0 ) = 0

f <( 2 ) = 60 x ( 2 ) 3 120 ( 2 ) = 240

f ( 2 ) = 60 x ( 2 ) 3 120 ( 2 ) = 240

Step 4: Draw conclusions based upon the Step 3 information.

i) If f ( critical value ) > 0 or positive, then you know that you have a local min at that critical value .

ii) If f ( critical value ) < 0 or negative, then you know that you have a local max at that critical value .

iii) If f ( critical value ) = 0 , then you know that your test failed , and you don’t know anything about your final answer. It still could be a local max, local min, but all you know is that the test failed and you wasted a bunch of your time.

Critical value : x = 0

f ( 0 ) = 0

Test Fails: You don’t know if it is a max or a min.

Critical value : x = 2

f ( 2 ) = 240

Since the 2 nd Derivative is negative, you know the shape is concave down. The only place a critical value can exist in a concave down shape is if it is a max.

You have a local max at x = 2 .

Local Max: f ( 2 ) = 3 ( 2 ) 5 20 ( 2 ) 3 = 64

Local Max: f ( 2 ) = 64

Local Max: ( 2 , 64 )

Critical value : x = 2

f ( 2 ) = 240

Since the 2 nd Derivative is positive, we know the shape is concave up. The only place a critical value can exist in a concave up shape is if it is a min.

You have a local min at x = 2 .

Local Min: f ( 2 ) = 3 ( 2 ) 5 20 ( 2 ) 3 = 64

Local Min: f ( 2 ) = 64

Local Min: ( 2 , 64 )

Final Result:

The function f ( x ) = 3 x 5 20 x 3 has a relative max of y = 64 at x = -2 .

The function f ( x ) = 3 x 5 20 x 3 has a relative min of y = -64 at x = 2 .

The 2 nd Derivative test failed to give us a conclusion about x = 0 .

This means you do not know if it is a max or a min. You only know that your test failed . If you needed to give a definitive answer, you would need to apply a 1 st Derivative Test. Looking at the graph you would see that in this situation that x = 0 is neither a max or a min, but that is not always the case.

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