Question

find the difference quotient \\(\frac{f(x+h)-f(x)}{h}\\), where \\(h \
eq 0\\), for the function \\(f(x) = 3x^2 - 12x\\). simplify your answer as much as possible.

Explanation:

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

Substitute \( x + h \) into \( f(x) = 3x^2 - 12x \):
\( f(x + h) = 3(x + h)^2 - 12(x + h) \)
Expand \( (x + h)^2 \) and distribute:
\( f(x + h) = 3(x^2 + 2xh + h^2) - 12x - 12h \)
\( f(x + h) = 3x^2 + 6xh + 3h^2 - 12x - 12h \)

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

Subtract \( f(x) = 3x^2 - 12x \) from \( f(x + h) \):
\( f(x + h) - f(x) = (3x^2 + 6xh + 3h^2 - 12x - 12h) - (3x^2 - 12x) \)
Simplify by canceling \( 3x^2 \) and \( -12x \):
\( f(x + h) - f(x) = 6xh + 3h^2 - 12h \)

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

eq 0 \) )
Divide \( f(x + h) - f(x) \) by \( h \):
\( \frac{f(x + h) - f(x)}{h} = \frac{6xh + 3h^2 - 12h}{h} \)
Factor out \( h \) in the numerator:
\( \frac{f(x + h) - f(x)}{h} = \frac{h(6x + 3h - 12)}{h} \)
Cancel \( h \) (since \( h
eq 0 \)):
\( \frac{f(x + h) - f(x)}{h} = 6x + 3h - 12 \)

Answer:

\( 6x + 3h - 12 \)