Question

find the difference quotient of f, that is, find \\( \frac { f ( x + h ) - f ( x ) } { h } \\), \\( h \
eq 0 \\), for the following function. \\( f ( x ) = x ^ { 2 } - 3 x + 9 \\) \\( \frac { f ( x + h ) - f ( x ) } { h } = \square \\) (simplify your answer.)

Explanation:

Step1: Find \( f(x + h) \)

Substitute \( x + h \) into \( f(x) \):
\( f(x + h)=(x + h)^2-3(x + h)+9 \)
Expand \( (x + h)^2 \) and \( -3(x + h) \):
\( (x + h)^2=x^2 + 2xh+h^2 \), \( -3(x + h)=-3x-3h \)
So \( f(x + h)=x^2 + 2xh+h^2-3x-3h + 9 \)

Step2: Calculate \( f(x + h)-f(x) \)

Substitute \( f(x + h) \) and \( f(x) \) into the expression:
\( f(x + h)-f(x)=(x^2 + 2xh+h^2-3x-3h + 9)-(x^2-3x + 9) \)
Remove the parentheses and combine like terms:
\( x^2 + 2xh+h^2-3x-3h + 9-x^2 + 3x-9=2xh+h^2-3h \)

Step3: Divide by \( h \) ( \( h

eq0 \))
\( \frac{f(x + h)-f(x)}{h}=\frac{2xh+h^2-3h}{h} \)
Factor out \( h \) from the numerator:
\( \frac{h(2x + h-3)}{h} \)
Cancel out \( h \) (since \( h
eq0 \)):
\( 2x + h-3 \)

Answer:

\( 2x + h-3 \)