# Derivative/Tangent Line Question

• Sep 17th 2007, 09:57 PM
Derivative/Tangent Line Question
1. Find an equation of the tangent line to the graph of f at the indicated point.

f(x) = 1/x+1
(0,1)

2. Find an equation of the line that is tangent to the graph of f and parallel to the given line.

f(x)=1/sqrtx, x + 2y-6=-

Any help would be greatly appreciated. Thanks!
• Sep 17th 2007, 11:42 PM
ticbol
Quote:

1. Find an equation of the tangent line to the graph of f at the indicated point.

f(x) = 1/x+1
(0,1)

2. Find an equation of the line that is tangent to the graph of f and parallel to the given line.

f(x)=1/sqrtx, x + 2y-6=-

Any help would be greatly appreciated. Thanks!

The slope of the tangent line is the value of the first derivative.

f(x) = 1/(x+1) = (x+1)^(-1)
f'(x) = -1(x+1)^(-2)
So at x=0,
m = f'(0) = -1(0+1)^(-2) = -1 /(1)^2 = -1
Using the point-slope form of the line, and the point (0,1):
(y -y1) = m(x -x1)
(y -1) = -1(x -0)
y -1 = -x
y = -x +1 ---------------the tangent line.

--------------------
If two lines are parallel, then they have the same slope.

f(x) = 1 /sqrt(x) = (x)^(-1/2)
f'(x) = (-1/2)x^(-3/2)
f'(x) = -1 /[2sqrt(x^3)] -------(i)

Slope of the line x + 2y-6=-
What?
Suppose the line is x + 2y-6 = 0
Transform it into the form y = mx +b,
x +2y -6 = 0
2y = -x +6
y = -(1/2)x +3
So, m = -1/2

Then, m = f'(x) = -1/2
We need the point of tangency so that we can use the point-slope form.

-1/2 = -1/[2sqrt(x^3)]
Divide both sides by (-1/2),
1 = 1/sqrt(x^3)
sqrt(x^3) = 1
Square both sides,
x^3 = 1
x = 1 ------at point of tangency.

What is the y there?
y = f(x) = 1/sqrt(x)
y = 1/sqrt(1)
y = 1
So, point of tangency is (1,1)
Then, with m = -1/2, and (1,1):
(y -1) = (-1/2)(x -1)
y = (-1/2)x +1/2 +1
y = -(1/2)x +3/2 ---------------the tangent line.