004 Sample Final A, Problem 11

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

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

Solution:

Step 1:
The difference quotient is ${\frac {f(x+h)-f(x)}{h}}={\frac {{\sqrt {x+h-3}}-{\sqrt {x-3}}}{h}}$.

Step 2:
Multiplying the numerator and denominator by ${\sqrt {x+h-3}}+{\sqrt {x-3}}$, we get
${\frac {f(x+h)-f(x)}{h}}={\frac {x+h-3-(x-3)}{h({\sqrt {x+h-3}}+{\sqrt {x-3}})}}$

Step 3:
Now, simplifying the numerator, we get
${\frac {f(x+h)-f(x)}{h}}={\frac {h}{h({\sqrt {x+h-3}}+{\sqrt {x-3}})}}$. Now, we can cancel the $h$ in the denominator.
Thus, ${\frac {f(x+h)-f(x)}{h}}={\frac {1}{({\sqrt {x+h-3}}+{\sqrt {x-3}})}}$.

Final Answer:
${\frac {f(x+h)-f(x)}{h}}={\frac {1}{({\sqrt {x+h-3}}+{\sqrt {x-3}})}}$

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