# 004 Sample Final A, Problem 11

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Find and simplify the difference quotient ${\displaystyle {\frac {f(x+h)-f(x)}{h}}}$ for ${\displaystyle f(x)={\sqrt {x-3}}}$

Foundations
1) ${\displaystyle f(x+h)=?}$
2) How do you eliminate the ${\displaystyle h}$ in the denominator?
1) We have ${\displaystyle f(x+h)={\sqrt {x+h-3}}}$
2) The difference quotient is ${\displaystyle {\frac {{\sqrt {x+h-3}}-{\sqrt {x-3}}}{h}}}$. To eliminate the ${\displaystyle h}$ in the denominator,
you need to multiply the numerator and denominator by ${\displaystyle {\sqrt {x+h-3}}+{\sqrt {x-3}}}$ (the conjugate).

Solution:

Step 1:
The difference quotient is ${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {{\sqrt {x+h-3}}-{\sqrt {x-3}}}{h}}}$.
Step 2:
Multiplying the numerator and denominator by ${\displaystyle {\sqrt {x+h-3}}+{\sqrt {x-3}}}$, we get
${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {x+h-3-(x-3)}{h({\sqrt {x+h-3}}+{\sqrt {x-3}})}}}$
Step 3:
Now, simplifying the numerator, we get
${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {h}{h({\sqrt {x+h-3}}+{\sqrt {x-3}})}}}$. Now, we can cancel the ${\displaystyle h}$ in the denominator.
Thus, ${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {1}{({\sqrt {x+h-3}}+{\sqrt {x-3}})}}}$.
${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {1}{({\sqrt {x+h-3}}+{\sqrt {x-3}})}}}$