IXL - Find absolute extrema on a closed interval (Calculus practice)

What are the absolute minimum and absolute maximum values of this function on the interval

[

–

]

(

)

Write exact, simplified answers.

Absolute minimum	Absolute maximum
Press the down arrow to access symbol buttons. Use arrow keys to navigate the symbol grid c c 𝜋 3 ln	Press the down arrow to access symbol buttons. Use arrow keys to navigate the symbol grid c 𝜋 3 ln

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question

What are the absolute minimum and absolute maximum values of this function on the interval

[

–

]

(

)

–

Write exact, simplified answers.

Absolute minimum	Absolute maximum
c	c

key idea

The Extreme Value Theorem states that if a function

is continuous on a closed interval

[

]

then

has both an absolute minimum and an absolute maximum on

[

]

Each extreme value occurs either at a critical point within the interval

(

)

or at one of the endpoints of

[

]

solution

The function

(

)

–

is continuous on the interval

[

–

]

So, each extreme value of

occurs either at a critical point within the interval

(

–

)

or at one of the endpoints of

[

–

]

First, find the

-values

where

has critical points. Take the derivative of

(

)

–

′

(

)

–

Notice that

′

(

)

is defined for all

-values

in the interval

(

–

)

So, to find the critical points, find the values of

in the interval where

′

(

)

′

(

)

–

(

–

)

(

–

)

(

)

3 or

–

Notice that

–

and

are not in the interval

(

–

)

So, on the interval

(

–

)

does not have any critical points.

Now, since there are no critical points on the interval

(

–

)

evaluate

at the endpoints of

[

–

]

Each extreme value of

must occur at one of these

-values.

Point type	x -value	Value of f ( x )
endpoint	– 2	f ( – 2 ) = 49
endpoint	3	f ( 3 ) = – 51

Finally, compare the values of

(

)

–

to find the absolute minimum and absolute maximum values on the interval

[

–

]

The absolute minimum value is

(

)

–

The absolute maximum value is

(

–

)

You can verify your solutions by looking at the graph of

(

)

–

on the interval

[

–

]

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