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

774,914 users

Find the set of x values where f(x) has relative minima and maxima

0 votes

A) f(x)=3x^2-15x-10

1) Find set of critical points of f(x)

2)find the intervals on which f(x) is increasing

3)find the intervals on which f(x) is decreasing

4) find the set of x-values where f(x) has relative minima

5) find the set of x-values where f(x) has relative maxima

B) f(x)= (x^2+2)/(2x^2+1)

1) Find set of critical points of f(x)

2)find the intervals on which f(x) is increasing

3)find the intervals on which f(x) is decreasing

4) find the set of x-values where f(x) has relative minima

5) find the set of x-values where f(x) has relative maxima

asked Sep 7, 2015 in CALCULUS by anonymous

6 Answers

0 votes

A-(1)

The function is image.

Differentiate with respect to x.

image

To find the critical points, equate image to zero.

image

Critical number is image.

Substitute image in the function.

image

Therefore, the critical point is image.

answered Sep 7, 2015 by Lucy Mentor
edited Sep 7, 2015 by Lucy
0 votes

A-(2)&(3)

The function is .

From A-(1), the critical number is .

The derivative function is image.

Consider test intervals as image and image.

Therefore the function is increasing in the interval image.

The function is decreasing in the interval image.

Solution:

A-(2): The function is increasing in the interval image.

A-(3): The function is decreasing in the interval image.

answered Sep 7, 2015 by Lucy Mentor
0 votes

A-(4)&(5)

The function is .

From A-(1), the critical number is .

The derivative function is image.

Differentiate with respect to x.

image

The sign of image is positive for all values of x.

Therefore, the function has relative minimum at image.

Relative minima is image.

No relative maxima.

Solution:

A-(4): Relative minima is image.

A-(5): No relative maxima.

answered Sep 7, 2015 by Lucy Mentor
0 votes

B-(1)

The function is image.

Differentiate with respect to image.

image

image.

To find the critical points, equate to zero.

image

Critical number is image.

Substitute image in the function.

image

Therefore, the critical point is image.

answered Sep 7, 2015 by cameron Mentor
edited Sep 7, 2015 by cameron
0 votes

B-(2)&(3)

The function is .

From B-(1), the critical point is .

The derivative function is .

Consider test intervals as and .

Therefore, the function is increasing in the interval .

The function is decreasing in the interval .

Solution:

B-(2): The function is increasing in the interval .

B-(3): The function is decreasing in the interval .

answered Sep 7, 2015 by cameron Mentor
edited Sep 7, 2015 by cameron
0 votes

B-(4)&(5)

The function is .

From B-(1), the critical number is .

The derivative function is image.

Differentiate with respect to x.

image

image.

Find the value of second derivative at .

image.

Therefore, the function has relative maximum at .

Relative maxima is image.

No relative minima.

Solution:

B-(4): Relative minima is image.

B-(5): No relative minima.

answered Sep 7, 2015 by cameron Mentor
edited Sep 7, 2015 by cameron

Related questions

...