Question

question
given the function $f(x) = 3x^2 - x$, express the value of $\frac{f(x+h)-f(x)}{h}$ in simplest form.

Explanation:

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

Substitute \( x + h \) into \( f(x) = 3x^2 - x \). So, \( f(x + h)=3(x + h)^2-(x + h) \). Expand \( (x + h)^2 \) using the formula \( (a + b)^2=a^2 + 2ab + b^2 \), we get \( f(x + h)=3(x^2+2xh + h^2)-x - h=3x^2+6xh + 3h^2-x - h \).

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

Subtract \( f(x)=3x^2 - x \) from \( f(x + h) \). So, \( f(x + h)-f(x)=(3x^2+6xh + 3h^2-x - h)-(3x^2 - x) \). Remove the parentheses: \( 3x^2+6xh + 3h^2-x - h - 3x^2 + x \). Combine like terms: \( 6xh+3h^2 - h \).

Step3: Divide by \( h \)

Divide \( f(x + h)-f(x) \) by \( h \), that is \( \frac{f(x + h)-f(x)}{h}=\frac{6xh + 3h^2 - h}{h} \). Factor out \( h \) from the numerator: \( \frac{h(6x + 3h - 1)}{h} \). Cancel out \( h \) (assuming \( h
eq0 \)): \( 6x + 3h - 1 \).

Answer:

\( 6x + 3h - 1 \)