find an equation of the tangent line to the hyperbola y=3/x?

Hi,

If f(x) = 3/x, then f(3 + h) - f(3) =

3/(3 + h) - 3/3 =

. 3
-------- - 1 = You used this in your question
3 + h

. 3 . . . .3 + h
-------- - ---------- =
3 + h . . 3 + h

3 - 3 - h
------------- =
3 + h

- h
-------- <==simplified ANSWER
3 + h

As h --> 0 this approaches 0.

I hope that helps!! :-)

The derivative of a function is
lim(h → 0) [f(x + h) - f(x)]/h.
We find the derivative to look for the slope of the tangent at (x, f(x)).

As f(x) = 3/x,
The derivative of f(x) is
lim(h → 0) [3/(x + h) - 3/x]/h.
To find the slope of the tangent at (3, 1), we substitute the x-value, 3, for x:
lim(h → 0) [3/(3 + h) - 3/3]/h
= lim(h → 0) [3/(3 + h) - 1]/h.

[3/(3 + h) - 1]/h
= [3 - (3 + h)]/(3 + h) * 1/h
= [-h]/(3 + h) * 1/h
= -1/(3 + h)

Thus, lim(h → 0) [3/(3 + h) - 1]/h
= lim(h → 0) -1/(3 + h)
= -1/(3 + 0)
= -1/3.

To find the c, substitute -1/3, 3 and 1 for m, x and y, respectively:
1 = -1/3 * 3 + c
c = 2.

Therefore, the the equation is y = -x/3 + 2.