Tangent line to the curve f(x)= x + 1/x

[theclock54]

Homework Statement

So I have to find the tangent line when x=5. I use the difference quotient, but I'm stuck at the algebra? I can't seem to figure out what steps I should take.

Here's what I have:

[(x+h)+1/(x+h)] - (x+1/x)

I've tried numerous steps on how to solve this, but can't do so. What I've basically have been trying is is to get the common denominators for the left and right sides, and then get another common denominator for both sides. What am I doing wrong?

Thank you in advance.

 

[micromass]

Did you see derivatives, yet?? Because that's what you need to use.

 

[gb7nash]

Homework Statement

So I have to find the tangent line when x=5. I use the difference quotient, but I'm stuck at the algebra? I can't seem to figure out what steps I should take.

Here's what I have:

[(x+h)+1/(x+h)] - (x+1/x)

Besides the limit, you're missing one thing. You want:

[tex]\frac{f(x+h)-f(x)}{h}[/tex]

So really, you have:

[tex]\frac{(x+h)+\frac{1}{x+h} - (x+\frac{1}{x})}{h}[/tex]

Simplify this a bit and on the *numerator part, find a common denominator like you were doing before.

 

[theclock54]

Yeah we're on derivatives, I just forgot to put the whole equation over h.

@gb7nash That's where I'm stuck, when find a common denominator and subtract both equations, the h doesn't cancel out. So I'm doing something wrong.

When I add the x+h and 1/x+h, I get [(x+h)^2 + 1]/(x+h) - (x^2+1)/x
Now do I get another common denominator?

 

[gb7nash]

Now do I get another common denominator?

Yes. Before you do anything with that, simplify the numerator. We can cancel something out.

 

[theclock54]

@gb7nash

Okay, so now I have: [x^2+2xh+h^2+1]/(x+h) - (x^2+1)/x

Get another com. denom then I have: [x(x^2+2xh+h^2+1) - (x^2+1)] / x(x+h)

 

[gb7nash]

I'm not quite sure what you're trying to do.

[tex]\frac{(x+h)+\frac{1}{x+h} - (x+\frac{1}{x})}{h} = \frac{h+\frac{1}{x+h} - \frac{1}{x}}{h}[/tex]

Now, looking at the numerator, we want to find a common denominator for [itex]\frac{1}{x+h}[/itex] and [itex]\frac{1}{x}[/itex]

How can we do this?

 

[theclock54]

Ohh, I was trying to add up all the fractions first. I didn't even notice the x's canceled out. Okay, so when I subtract 1/x+h - 1/x I get -h/x+h

After the mess, I get my answer to be (h+x-1)/x+h