equation of line tangent to f(x)= 1/sqrt x^2 - 3 at x=2

[jessica716]

Hi I was hoping someone could help me with this problem

f(x)= 1/sqrt x^2 - 3 at x=2

I know I have to take the derivative then replace x with 2 but I am not getting the answer that I am supposed to. any help would be appreciated.

Thanks

 

[o_O]

What's f(x)? \(\displaystyle \frac{1}{\sqrt{x^{2} - 3}\)?

Just imagine \(\displaystyle f(x) = \left((x^{2} - 3)^{1/2}\right)^{-1}\), simplify, and use your chain rule.

 

[jessica716]

I keep getting 1 when i replace x with 2 and then do the point slope formula to get the equation and the equation doesn't match the choices i have for the answer.

 

[o_O]

I keep getting 1 with i replace x with 2

Umm ... you might have to repeat that again

and then do the point slope formula to get the equation and the equation doesn't match the choices i have for the answer.

What's the answer you got?

 

[jessica716]

the equation I got was y=x-1 my choices however are (1). y=-2x+3, (2). y=-2x (3). y=2x+3, (4) y=-2x+5.

I have been working on this problem for an hour now and I can't figure out what I am doing wrong. I really appreciate your help.

 

[o_O]

Oh whoops didn't read your question clearly. Well, either way, what did you get for your derivative? We have to get past this step first.

Actually, show us what you've done and we'll help you from there.

 

[jessica716]

for the derivative i got -1(1/2(2x)^1/2)^-2 and then when i replaced x with 2 I get 1. I also tried using the quotient rule and got 1x^1/2 /x^2-3 and when I replaced x with 2 i got 1 again. Then when i use the point slope formula with the point (2,1) i don't get one of the options listed.

 

[o_O]

How did you get your derivative? Your function is \(\displaystyle f(x) = \frac{1}{\sqrt{x^{2} - 3}}\) right?

\(\displaystyle f(x) = \frac{1}{\sqrt{x^{2} - 3}} = (x^{2} - 3)^{-1/2})\)

Use your chain rule and show us your steps so we can see if any mistakes occur.

 

[jessica716]

would the derivative be : -1/2(2x)^1/2? That's what I keep getting.

 

[o_O]

Not quite. Recall the chain rule:
\(\displaystyle [f(g(x))]' = f'(g(x)) \cdot g'(x)\)

So if we apply this your function:
Let f(x) = x^-1/2 and g(x) = x² - 3. Then:

\(\displaystyle [f(g(x))]' = [g(x)^{-1/2}]' \cdot g'(x)\)

\(\displaystyle = -\frac{1}{2}g(x) \cdot g'(x)\)

\(\displaystyle = -\frac{1}{2}(x^{2} - 3)^{-3/2} \cdot (x^{2} - 3)'\)

etc. etc.

 

[jessica716]

ok, so would the derivative be -1/2(x^2-3)^1/2 * (2x) ?

never mind, when I replace x with 2 i still get one. Thanks for your help, i just can't figure this out.

 

[o_O]

Well, your exponent isn;t quite right there but other than that, f'(2) should work out to be -2 when you plug in 2.

 

[jessica716]

can you tell me what my exponent should be, i can't seem to figure it out and it's driving me crazy.

 

[o_O]

I wrote this earlier:
\(\displaystyle -\frac{1}{2}(x^{2} - 3)^{-3/2} \cdot (x^{2} - 3)'\)

Can you see the exponent now?

 

[jessica716]

i think i get it. Thanks so much