IXL - Find local extrema using the first derivative test (Calculus practice)

Use the first derivative test to find the local extrema of

(

)

List the

-values

of the extrema.

(

)

–

Write one exact simplified answer in each box. You can add or remove boxes. If

has no local extrema, remove all boxes.

-values

of local minima:

-values

of local maxima:

ref_doc_title.

Learn with an example

question

Use the first derivative test to find the local extrema of

(

)

List the

-values

of the extrema.

(

)

–

Write one exact simplified answer in each box. You can add or remove boxes. If

has no local extrema, remove all boxes.

-values

of local minima:

-values

of local maxima:

key idea

A function

has a critical point at

in its domain if either

′

(

)

′

(

)

does not exist. If

has local extrema, then each must occur at a critical point.

The first derivative test states that if

is continuous on an open interval that contains a critical point at

and is differentiable on the interval except possibly at

then the following are true:

If
f
′
changes sign from negative to positive at
c
,
then
f
has a local minimum at
x
=
c
.
If
f
′
changes sign from positive to negative at
c
,
then
f
has a local maximum at
x
=
c
.
If
f
′
does not change sign at
c
,
then
f
does not have a local extremum at
x
=
c
.

solution

You were asked to find the

-values

where

(

)

–

has local extrema. To do so, you can use the first derivative test.

First, find the

-values

where

has critical points. Take the derivative of

(

)

–

′

(

)

–

Notice that

′

(

)

is defined for all

-values.

So, to find the critical points, find the

-values

where

′

(

)

′

(

)

–

(

–

)

–

(

–

)

(

)

3 or

–

So,

has critical points at

–

and

These are the only possible

-values

where

can have local extrema.

Now, since

(

)

–

is continuous and differentiable on the interval

(

–

∞

)

you can use the first derivative test to determine whether each critical point you found is at a local minimum, a local maximum, or neither.

To do so, break the interval

(

–

∞

)

into subintervals using the

-values

of the critical points. Then, determine whether

′

is positive or negative on each subinterval.

Notice that the derivative factors nicely as

′

(

)

–

(

–

)

(

)

So, you can determine whether

′

is positive or negative on each subinterval using a sign chart. For each subinterval, determine the sign of each factor. The sign of

′

(

)

will be the product of those signs. Note that you could also determine the sign of

′

(

)

by choosing test values within each interval.

	– 3	x – 3	x + 3	Sign of f ′ ( x )
( – ∞ , – 3 )	—	—	—	—
( – 3 , 3 )	—	—	+	+
( 3 , ∞ )	—	+	+	—

Now, use the first derivative test by analyzing how the sign of

′

changes at each critical point:

f
has a local minimum at
x
=
–
3
since
f
′
changes from negative to positive at that
x
-value.
f
has a local maximum at
x
=
3
since
f
′
changes from positive to negative at that
x
-value.

You can verify your solutions by looking at the graph of

(

)

–

$Confetti$

Excellent!

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