Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

    
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

776,759 users

help? derivative-trig

0 votes

asked May 1, 2015 in CALCULUS by anonymous

2 Answers

0 votes

(7)

Step 1:

The function is image and image.

Slope of the tangent line equation is derivative of the function.

image.

Differentiate on each side with respect to .

image.

Product rule of derivatives : .

image

Substitute image in derivative function.

image

Slope of the tangent line equation is image.

Step 2:

Find the point of tangency.

Substitute image in the function.

image

Tangent point is image.

Find the equation of tangent line.

Point slope form of the line equation : .

Substitute image and image in point slope form of line equation.

image

Tangent line equation is image.

Solution :

Tangent line equation is image.

answered May 1, 2015 by cameron Mentor
edited May 1, 2015 by cameron
0 votes

(8)

Step 1:

The function is .

Differentiate on each side with respect to .

.

Find the critical points.

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

Find the critical point by graphing the derivative function.

Graph:

Graph the derivative function .

Observe the graph:

Locate the values of x where the value of the derivative function is zero.

Critical points are and .

answered May 1, 2015 by cameron Mentor
edited May 1, 2015 by cameron

Continued...

Step 2:

The range is .

The test intervals are , , , and .

Therefore the function is increasing on the intervals , and .

The function is decreasing on the intervals and .

Solution:

The function is increasing on the intervals , and  .

The function is decreasing on the intervals and  .

Related questions

asked May 3, 2015 in CALCULUS by anonymous
asked Nov 3, 2014 in CALCULUS by anonymous
asked Oct 25, 2014 in CALCULUS by anonymous
asked Nov 18, 2014 in PRECALCULUS by anonymous
asked Oct 26, 2014 in CALCULUS by anonymous
...