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help-derivatives-[shape of]????

0 votes

asked May 6, 2015 in CALCULUS by anonymous

6 Answers

0 votes

(12)

Step 1:

Thu function is .

Differentiate on each side with respect to .

.

Step 2:

Find the critical points.

A critical number of a function is a number in the domain of such that either or does not exist.

Since is a polynomial it is continuous for all values of .

Thus, the critical points exist when image.

Equate image to zero.

and

and

The critical point are and .

Solution:

The critical point are and .

answered May 6, 2015 by Lucy Mentor
0 votes

(13)

Step 1:

Thu function is .

The critical point are and .             (From(12))

The test intervals are , and .

The function is increasing on the intervals and .

The function is decreasing on the interval .

Solution :

The function is increasing on the intervals and .

The function is decreasing on the interval .

answered May 6, 2015 by Lucy Mentor
0 votes

(14)

Step 1:

Find the local maximum and local minimum.

The function has a local maximum at , because image changes its sign from positve to negative.

Substitute in .

Local maximum is .

The function has a local minimum at , because image changes its sign from negative to positive.

Substitute in .

Local minimum is .

Solution:

Local maximum is .

Local minimum is .

answered May 6, 2015 by Lucy Mentor
0 votes

(15)

Step 1:

Thu function is .

.

Differentiate image on each side with respect to .

.

Equate to zero.

Consider the test intervals as and .

The graph of the function is concave up on the interval .

The graph of the function is concave down on the interval .

Solution:

The function is concave up on the interval and concave down on the interval .

answered May 6, 2015 by Lucy Mentor
0 votes

(16)

Step 1:

Thu function is .

Find the inflection points.

Inflection Point :

Inflection point is a point on the curve at which the function changes from concave up to down or vice versa.

The curve changes concave down to concave up at .           (from (15)).

Substitute in the function.

Inflection point is .

Solution :

Inflection point is .

answered May 6, 2015 by Lucy Mentor
0 votes

(17)

Thu function is .

The function is increasing on the intervals and .

The function is decreasing on the interval .

Local maximum is .

Local minimum is .

The function is concave up on the interval and concave down on the interval .

Inflection point is .

Graph :

Graph the function with the above characteristics.

answered May 6, 2015 by Lucy Mentor

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