Question

for the function $f(x) = 4x^2 + 5x$, evaluate and simplify.\
$\frac{f(x + h) - f(x)}{h} = \square$\
question help: \
video \
written example

Explanation:

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

Substitute \( x + h \) into \( f(x) = 4x^2 + 5x \):
\( f(x + h) = 4(x + h)^2 + 5(x + h) \)
Expand \( (x + h)^2 \): \( (x + h)^2 = x^2 + 2xh + h^2 \)
So, \( f(x + h) = 4(x^2 + 2xh + h^2) + 5x + 5h = 4x^2 + 8xh + 4h^2 + 5x + 5h \)

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

Subtract \( f(x) = 4x^2 + 5x \) from \( f(x + h) \):
\( f(x + h) - f(x) = (4x^2 + 8xh + 4h^2 + 5x + 5h) - (4x^2 + 5x) \)
Simplify by canceling \( 4x^2 \) and \( 5x \):
\( f(x + h) - f(x) = 8xh + 4h^2 + 5h \)

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

eq 0 \) )
Divide \( f(x + h) - f(x) \) by \( h \):
\( \frac{f(x + h) - f(x)}{h} = \frac{8xh + 4h^2 + 5h}{h} \)
Factor out \( h \) in the numerator:
\( \frac{h(8x + 4h + 5)}{h} \)
Cancel \( h \) (since \( h
eq 0 \)):
\( 8x + 4h + 5 \)

Answer:

\( 8x + 4h + 5 \)