equation of a tangent

• Oct 3rd 2010, 04:54 PM
nightrider456
equation of a tangent
This is probably pretty easy but i am confused with how to continue

The problem: Find an equation of the straight line having a slope $\frac{1}{4}$ that is tangent to the curve $y=\sqrt{x}$

I pluged some values in to a formula my prof. gave me $\frac{f(x_{1}+h)-f(x_{1})}{h}$

Here is what I have so far:
$\frac{1}{4}=\lim_{h\rightarrow0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h}$
But i don't know where to go from here
Thanks for looking
• Oct 3rd 2010, 05:10 PM
skeeter
Quote:

Originally Posted by nightrider456
This is probably pretty easy but i am confused with how to continue

The problem: Find an equation of the straight line having a slope $\frac{1}{4}$ that is tangent to the curve $y=\sqrt{x}$

I pluged some values in to a formula my prof. gave me $\frac{f(x_{1}+h)-f(x_{1})}{h}$

Here is what I have so far:
$\frac{1}{4}=\lim_{h\rightarrow0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h}$
But i don't know where to go from here
Thanks for looking

$\displaystyle \frac{1}{4} = \lim_{h\to 0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h} \cdot \frac{\sqrt{x_{1}+h}+\sqrt{x_{1}}}{\sqrt{x_{1}+h}+ \sqrt{x_{1}}}$

ring any bells?
• Oct 3rd 2010, 08:27 PM
nightrider456
So I would then simplify from there?
• Oct 4th 2010, 05:57 AM
skeeter
Quote:

Originally Posted by nightrider456
So I would then simplify from there?

I gave you the first step in finding the limit on the right hand side of the equation ... finish it by finding that limit, set it equal to 1/4 and solve for $x_1$.
• Oct 4th 2010, 09:03 AM
HallsofIvy
Are you required to use the definition of limit? If you know the "power rule", that the derivative of $x^r$, where r is a constant, is $rx^{r-1}$, then you can use that with $r= \frac{1}{2}$.