# equation of the tangent line

• Oct 16th 2007, 07:31 AM
leviathanwave
equation of the tangent line
Hi, I'm a little stuck on this problem. I'm not sure if I get the derivative and then use y=mx+b or what, any help much appreciated.

Find the equation of the line tangent to the graph of the given function at the point with the indicated x coordinate.

f(x) = x/(x^2 + 1); x=1
• Oct 16th 2007, 07:38 AM
Jhevon
Quote:

Originally Posted by leviathanwave
Hi, I'm a little stuck on this problem. I'm not sure if I get the derivative and then use y=mx+b or what, any help much appreciated.

i think it's nicer to use \$\displaystyle y - y_1 = m(x - x_1)\$, but if you must use y = mx + b, that's fine.

Quote:

Find the equation of the line tangent to the graph of the given function at the point with the indicated x coordinate.

f(x) = x/(x^2 + 1); x=1
you start by finding the derivative of f(x), this gives a formula for the slope at any x. this will give you the value for m.

afterwards, you need to find corresponding x and y values. you were already given the x-value. to find the y-value, plug in the given x-value into the original equation and solve for y (which is the same as f(x)). then plug in all those values into y = mx + b. b will be your only unknown, solve for it. then rewrite the equation y = mx + b, filling in only m and b, that will be the tangent line

try it and see what you get
• Oct 16th 2007, 01:45 PM
leviathanwave
so y = 1/2?
• Oct 16th 2007, 01:47 PM
Jhevon
Quote:

Originally Posted by leviathanwave
so y = 1/2?

yes, when x = 1, y = 1/2. so the point is (1, 1/2)

now continue