find the slope of the secant line for the function f(x)=3.2x^2-1.6x passing thro

Question

find the slope of the secant line for the function f(x)=3.2x^2-1.6x passing through the points where x=1 and x= 0.8. Round the answer to two decimal places.

Approximate the slpe of the line tangent to f(x)= 3x^2 at x=2. Be sure to use anincrement that is small enough for the answer to be accrate to two decimal places.

The point P(1,1)lies on the curve f(x)=x^4. Let Q(xo,xo^4) be an arbitrary point on the curve, and let s be the secant line drawn through P and Q. Let the slope of the secant line drawn through P and Q be denoted by ms. Find lim ( x0 approaches 1) m s.

Find the slope of the seant line for he fuction f(x)=2x^2+1 passing through the points (1,3) and (1.1, 3.42). Round your answer to one decimal places

Explanation / Answer

If you learned how to derive, derive the equation for tangent questions.

1. At x=0.8, f(x) = 0.768

At x = 1, f(x) = 1.6

slope = rise/run = (1.6-.768)/(1-0.8)

= 4.16

2. If you learned how to derive, derive f(x) to get f'(x)

f'(x) = 6x

f'(2) = 12.00

If not, use secant and limit.

f(2) = 12

f(x) = 3x^2

slope of secant = rise/run = (3x^2-12)/(x-2)

slope of tangent = lim (x -> 2) (3x^2-12)/(x-2)

= lim (x -> 2) 3(x^2-4)/(x-2)

= lim (x -> 2) 3(x+2)(x-2)/(x-2)

= lim (x -> 2) 3(x+2)

= 12

3.

slope of secant = rise/run = (xo^4-1)/(xo-1)

slope of tangent = lim (xo -> 1) (xo^4-1)/(xo-1)

= lim (xo -> 1) (xo^2+1)(xo+1)(xo-1)/(xo-1)

= lim (xo -> 1) (xo^2+1)(xo+1)

= 4

4. Simply find secant slope

slope = rise/run = (3.42-3)(1.1-1)

= 0.42-0.1

=4.2

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