Question

for the function $f(x) = 3x^2 - 2x$, evaluate and simplify.\\(\frac{f(x + h) - f(x)}{h} = \boxed{}\\)\
question help: \\(\boxed{}\\) video

Explanation:

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

Substitute \( x + h \) into \( f(x) = 3x^2 - 2x \):
\( f(x + h) = 3(x + h)^2 - 2(x + h) \)
Expand \( (x + h)^2 \) using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):
\( f(x + h) = 3(x^2 + 2xh + h^2) - 2x - 2h \)
Distribute the 3 and -2:
\( f(x + h) = 3x^2 + 6xh + 3h^2 - 2x - 2h \)

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

Subtract \( f(x) = 3x^2 - 2x \) from \( f(x + h) \):
\( f(x + h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x) \)
Remove the parentheses and combine like terms:
\( 3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x = 6xh + 3h^2 - 2h \)

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

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

Answer:

\( 6x + 3h - 2 \)