Question

for the function ( f(x)=2x^2+3x ): (a) a simplified form of the difference quotient ( \frac{f(x+h)-f(x)}{h} ), when ( h
eq 0 ), is

Explanation:

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

Substitute \( x + h \) into \( f(x) = 2x^2 + 3x \).
\( f(x + h)=2(x + h)^2+3(x + h) \)
Expand \( (x + h)^2 \): \( (x + h)^2=x^2 + 2xh+h^2 \)
So \( f(x + h)=2(x^2 + 2xh + h^2)+3x + 3h=2x^2+4xh + 2h^2+3x + 3h \)

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: \( 2x^2+4xh + 2h^2+3x + 3h - 2x^2-3x = 4xh+2h^2 + 3h \)

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

eq0 \))
Compute \( \frac{f(x + h)-f(x)}{h}=\frac{4xh + 2h^2+3h}{h} \)
Factor out \( h \) in the numerator: \( \frac{h(4x + 2h + 3)}{h} \)
Cancel out \( h \) (since \( h
eq0 \)): \( 4x + 2h+3 \)

Answer:

\( 4x + 2h + 3 \)