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.

13,459 questions

17,854 answers

1,446 comments

785,590 users

please help-need help asap

0 votes

2)  Use the information from #1 to estimate the slope of the tangent line to f(x) at x = 3 and write the equation of the tangent line.

asked Mar 29, 2015 in CALCULUS by anonymous

9 Answers

0 votes

a)

f(x) = x^3 - 3x^2 

At  x = 3:

f(3) = 3^3 - 3(3)^2

       = 27 - 27

       = 0

Point P is ( 3 , 0)

f(x) = x^3 - 3x^2 

At  x = 3.5:

f(3) = 3.5^3 - 3(3.5)^2

       = 6.125

Point Q is ( 3.5 , 6.125)

Find the slope of PQ:

m  = (y2 - y1)/(x2 - x1)

     = (6.125 - 0)/(3.5 - 3)

     = 12.25

answered Mar 29, 2015 by bradely Mentor
0 votes

Step 1:

(b)

The function is .

At the point , .

Find -coordinate value at .

Substitute in .

The point is .

Step 2:

At the point , .

Find  corresponding -coordinate value at .

.

Substitute in .

The point is .

Step 3:

Find the slope of the .

Slope of the two points  and is .

Substitute and in slope formula.

Slope of the secant line is .

Solution:

(b)  Slope of the secant line is .

answered Mar 30, 2015 by cameron Mentor
edited Mar 30, 2015 by cameron
0 votes

Step 1:

(c)

The function is .

At the point , .

Find  corresponding -coordinate value at .

.

Substitute in .

The point is .

Step 2:

Find the slope of the .

Slope of the two points  and is .

Substitute and in above formula.

Slope of the secant line is .

Solution:

Slope of the secant line is .

answered Mar 30, 2015 by cameron Mentor
edited Mar 30, 2015 by cameron
0 votes

Step 1:

(d) The function is .

At the point , .

Find  corresponding -coordinate value at .

.

substitute in .

The point is .

Step 2:

Find the slope of the .

Slope of the two points  and is .

Substitute and in above formula.

Slope of the secant line is .

Solution:

Slope of the secant line is .

answered Mar 30, 2015 by cameron Mentor
0 votes

Step 1:

(e)

The function is .

At the point , .

Find the corresponding -coordinate value at .

.

substitute in .

The point is .

Step 2:

Find the slope of the .

Slope of the two points  and is .

Substitute and in above formula.

Slope of the secant line is .

Solution:

Slope of the secant line is .

answered Mar 30, 2015 by cameron Mentor
edited Mar 30, 2015 by cameron
0 votes

Step 1:

(f)

The function is .

At the point , image.

Find the corresponding -coordinate value at image.

.

substitute image in .

image

The point is image.

Step 2:

Find the slope of the :

Slope of the two points  and is .

Substitute and image in above formula.

image

Slope of the secant line is image.

Solution:

Slope of the secant line is image.

answered Mar 30, 2015 by cameron Mentor
0 votes

Step 1:

(g)

The function is .

At the point , image.

Find  corresponding -coordinate value at image.

.

Substitute image in .

image

The point is image.

Step 2:

Find the slope of the :

Slope of the two points  and is .

Substitute and image in above formula.

image

Slope of the secant line is image.

Solution:

Slope of the secant line is image.

answered Mar 30, 2015 by cameron Mentor
0 votes

Step 1:

(h) The function is .

At the point , image.

Find  corresponding -coordinate value at image.

.

substitute image in .

image

The point is image.

Step 2:

Find the slope of the :

Slope of the two points  and is .

Substitute and image in above formula.

image

Slope of the secant line is image.

Solution:

Slope of the secant line is image.

answered Mar 30, 2015 by cameron Mentor
0 votes

Step 1:

(2)

The function is .

Slope of the tangent is derivative of the function.

Differentiate on each side with respect to image.

image

Slope of the tangent line at image.

image

Slope of the tangent line is image.

Step 2:

Point image is .

Point-slope form of line equation: .

Substitute slope image and the point image in the above formula.

image

The equation of a tangent line image.

Solution:

The equation of a tangent line image.

answered Mar 30, 2015 by cameron Mentor
edited Mar 30, 2015 by cameron

Related questions

asked Feb 14, 2015 in CALCULUS by anonymous
asked Nov 4, 2014 in CALCULUS by anonymous
asked Sep 12, 2014 in CALCULUS by anonymous
asked Apr 27, 2015 in CALCULUS by anonymous
asked Mar 3, 2015 in CALCULUS by anonymous
...