004 Sample Final A, Problem 11

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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?
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, ${\frac {f(x+h)-f(x)}{h}}={\frac {1}{({\sqrt {x+h-3}}+{\sqrt {x-3}})}}$ .
${\frac {f(x+h)-f(x)}{h}}={\frac {1}{({\sqrt {x+h-3}}+{\sqrt {x-3}})}}$ 