Question

find the difference quotient of f, that is, find \\( \frac { f ( x + h ) - f ( x ) } { h } \\), \\( h \
eq 0 \\), for the following function. be sure to simplify. \\( f ( x ) = x ^ { 2 } - 6 x + 5 \\) \\( \frac { f ( x + h ) - f ( x ) } { h } = \\) (simplify your answer.)

Explanation:

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

Substitute \(x + h\) into \(f(x)\):
\(f(x + h)=(x + h)^{2}-6(x + h)+5\)
Expand \((x + h)^{2}\) and \(-6(x + h)\):
\((x + h)^{2}=x^{2}+2xh + h^{2}\), \(-6(x + h)=-6x-6h\)
So \(f(x + h)=x^{2}+2xh + h^{2}-6x-6h + 5\)

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

Subtract \(f(x)=x^{2}-6x + 5\) from \(f(x + h)\):
\(f(x + h)-f(x)=(x^{2}+2xh + h^{2}-6x-6h + 5)-(x^{2}-6x + 5)\)
Remove the parentheses and combine like terms:
\(x^{2}+2xh + h^{2}-6x-6h + 5 - x^{2}+6x - 5=2xh + h^{2}-6h\)

Step3: Calculate the difference quotient \(\frac{f(x + h)-f(x)}{h}\)

Divide \(f(x + h)-f(x)=2xh + h^{2}-6h\) by \(h\) (\(h
eq0\)):
\(\frac{2xh + h^{2}-6h}{h}=\frac{h(2x + h - 6)}{h}\)
Cancel out \(h\) (since \(h
eq0\)):
\(2x + h - 6\)

Answer:

\(2x + h - 6\)