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

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

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 .

edited Mar 30, 2015 by cameron

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 .

edited Mar 30, 2015 by cameron

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 .

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 .

edited Mar 30, 2015 by cameron

Step 1:

(f)

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 .

Step 1:

(g)

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 .

Step 1:

(h) 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 .

Step 1:

(2)

The function is .

Slope of the tangent is derivative of the function.

Differentiate on each side with respect to .

Slope of the tangent line at .

Slope of the tangent line is .

Step 2:

Point is .

Point-slope form of line equation: .

Substitute slope and the point  in the above formula.

The equation of a tangent line .

Solution:

The equation of a tangent line .