Question

chapter 1.1

1. given $f(x) = 2x^2 - 3x$. evaluate and simplify $dfrac{f(x+h)-f(x)}{h}$

Explanation:

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

Substitute \( x + h \) into \( f(x) = 2x^2 - 3x \). So, \( f(x + h)=2(x + h)^2-3(x + h) \). Expand \( (x + h)^2 \) as \( x^2 + 2xh+h^2 \), then \( f(x + h)=2(x^2 + 2xh + h^2)-3x - 3h=2x^2+4xh + 2h^2-3x - 3h \).

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

Subtract \( f(x)=2x^2 - 3x \) from \( f(x + h) \). \( f(x + h)-f(x)=(2x^2+4xh + 2h^2-3x - 3h)-(2x^2 - 3x) \). Simplify by canceling \( 2x^2 \) and \( - 3x \), we get \( 4xh+2h^2 - 3h \).

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

eq0 \))
Divide \( f(x + h)-f(x) \) by \( h \): \( \frac{f(x + h)-f(x)}{h}=\frac{4xh + 2h^2-3h}{h} \). Factor out \( h \) from the numerator: \( \frac{h(4x + 2h-3)}{h} \). Cancel \( h \) (since \( h
eq0 \)), we get \( 4x + 2h-3 \).

Answer:

\( 4x + 2h - 3 \)