Question

Can anyone help me work through this precalc problem?
Find the slope of the tangent line to the graph of f(x) = (1/x) at the point (1,1)

Answer 1

To find the slope of the tangent line at a point, find the value of the derivative at that point.

Answer 2

I get mixed up in the simplification.
I've been able to do the others, but the fraction ones throw me everytime.

Answer 3

Rewrite the function as \(f(x)=x^{-1}\) if that makes it easier.

Answer 4

I don't think I am doing this right. I get to (1/1+h) -1/h and I get stuck.
I think that's where I go wrong each time.

Answer 5

Are you still using the limit definition of the derivative?

Answer 6

how is this a precalc problem? you'll need to take the derivative...
unless you plan on finding the slope of the tangent line at x=1 using the limit process...

Answer 7

Sorry, hadn't realized. Give me a minute, I'll type something up.

Answer 8

that's what they want me to do. Use the limiting process to find the slope.

Answer 9

\[
\large\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to0}\frac{\frac{-h}{x(x+h)}}{h}=\lim_{h\to0}\frac{-1}{x(x+h)}
\]Can you take it from there?

Answer 10

I think so. Any tips about how to convert this to the limiting process my text wants
us to use?

Answer 11

I don't know what your text wants you to use, sorry. This is a limiting process. Does your text want you to graph secant lines or something?

Answer 12

Sorry. It wants me to plug in the values for x at the given point. So (1/x+h) becomes (1/1+h)

Answer 13

\[\large\lim_{h\to0}\frac{\frac{1}{1+h}-\frac{1}{1}}{h}=\lim_{h\to0}\frac{\frac{1-(1+h)}{1(1+h)}}{h}=\lim_{h\to0}\frac{\frac{-h}{1(1+h)}}{h}=\lim_{h\to0}\frac{-1}{1(1+h)}\]

Answer 14

Okay. thanks!

Answer 15

:)