# Math Help - Calculas question

1. ## Calculas question

I hope this is the right area to post this question.

Find the equation of the tangent to the curve y=2/x+1 at the point where x=3.

y=2/x+1
y=2/4
=1/2 (3,1/2)

2(x+1)^-1
-2(x+1)^-2
-2/(x+1)^2
2/(3+1)(3+1)
2/8
=1/4
therefore in y=mx+b form:
y=1/4x+b
1/2=1/4(3)+b
b=2/3
y=1/4x+2/3

I have done that and I think I have done it correctly but the answer I have is
x+8y-7=0
How can I get this answer or is this wrong.

2. Originally Posted by Awsom Guy
I hope this is the right area to post this question.

Find the equation of the tangent to the curve y=2/x+1 at the point where x=3.

y=2/x+1 I assume you mean 2/(x + 1)
y=2/4
=1/2 (3,1/2)

2(x+1)^-1
-2(x+1)^-2
-2/(x+1)^2
2/(3+1)(3+1) Correct up to here. But remember that 4x4 = 16
...

3. ## I still dont get it.

okay I will re-do that last bit.
2/16
1/8
y=mx+b
y=1/8x+b
1/2=1/8(3)+b
2/3=b

therefore:
y=1/8x+1/1/3

but I still don't get the right answer of x+8y-7=0
Thanks

4. Originally Posted by Awsom Guy
okay I will re-do that last bit.
2/16
1/8
y=mx+b
y=1/8x+b
1/2=1/8(3)+b Correct up to here.

1/2 = 3/8 + b

b = 1/2 - 3/8

b = 4/8 - 3/8

b = 1/8.

So y = (1/8)x + 1/8

8y = x + 1.

I disagree with the answer you have been given. Did you copy it down correctly?

5. ## Thanks

Originally Posted by Prove It
I disagree with the answer you have been given. Did you copy it down correctly?
[/COLOR][/COLOR]
Thanks for that.
The answer was in a maths booklet I have the exact question with me. I am not sure why the answer is wrong,I'll make sure I recheck everything.
Thanks

6. Actually, I figured out what's wrong.

You have $\frac{dy}{dx} = -\frac{2}{(x + 1)^2}$.

So at $x = 3$ you have

$\frac{dy}{dx} = -\frac{2}{4^2} = -\frac{2}{16} = -\frac{1}{8}$, not $\frac{1}{8}$.

The rest should follow.