# Calculus Problem

• Mar 12th 2012, 01:22 AM
BobBali
Calculus Problem
Hi All,

Question asks to find the slope of tangent from First Principles for $\displaystyle f(x) = \sqrt{x-6}$ at point of contact x= 10
First find:

$\displaystyle f(10)= 2$

Then using Slope-Limit equation: $\displaystyle f(x) = \frac{f(x) - f(a)}{x - a}$

$\displaystyle \frac{\sqrt{x+10} - \sqrt{6} - 2}{x - 10}$

$\displaystyle \frac{\sqrt{x+10} - \sqrt{6} -2}{x - 10}$

$\displaystyle \frac{\sqrt{x+10} - \sqrt{6} -2}{\sqrt{x} + \sqrt{10} - \sqrt{x} - \sqrt{-10}$

After this i am at a loss...
• Mar 12th 2012, 02:09 AM
princeps
Re: Calculus Problem
Quote:

Originally Posted by BobBali
Hi All,

Question asks to find the slope of tangent from First Principles for $\displaystyle f(x) = \sqrt{x-6}$ at point of contact x= 10
First find:

$\displaystyle f(10)= 2$

Then using Slope-Limit equation: $\displaystyle f(x) = \frac{f(x) - f(a)}{x - a}$

$\displaystyle \frac{\sqrt{x+10} - \sqrt{6} - 2}{x - 10}$

$\displaystyle \frac{\sqrt{x+10} - \sqrt{6} -2}{x - 10}$

$\displaystyle \frac{\sqrt{x+10} - \sqrt{6} -2}{\sqrt{x} + \sqrt{10} - \sqrt{x} - \sqrt{-10}$

After this i am at a loss...

$\displaystyle m= \displaystyle \lim_{x \to 10} \frac{f(x)-f(a)}{x-a}$

$\displaystyle m= \displaystyle \lim_{x \to 10} \frac{\sqrt{x-6}-2}{x-10} \cdot \frac {\sqrt{x-6}+2}{\sqrt{x-6}+2}$

$\displaystyle m=\displaystyle \lim_{x \to 10} \frac{x-10}{(x-10)(\sqrt{x-6}+2)}=\displaystyle \lim_{x \to 10} \frac{1}{(\sqrt{x-6}+2)}=\frac{1}{4}$