Question

find and simplify the difference quotient \\(\frac{f(x + h) - f(x)}{h}\\), \\(h \
eq 0\\) for the given function.\\(f(x) = -2x^2 - x - 4\\)

Explanation:

Step1: Compute $f(x+h)$

$f(x+h) = -2(x+h)^2 - (x+h) - 4$
Expand $(x+h)^2$:
$f(x+h) = -2(x^2 + 2xh + h^2) - x - h - 4 = -2x^2 -4xh -2h^2 -x -h -4$

Step2: Calculate $f(x+h)-f(x)$

Substitute $f(x) = -2x^2 -x -4$:
$f(x+h)-f(x) = (-2x^2 -4xh -2h^2 -x -h -4) - (-2x^2 -x -4)$
Simplify by canceling like terms:
$f(x+h)-f(x) = -4xh -2h^2 -h$

Step3: Divide by $h$ (h≠0)

$\frac{f(x+h)-f(x)}{h} = \frac{-4xh -2h^2 -h}{h}$
Factor out $h$ in numerator:
$\frac{f(x+h)-f(x)}{h} = \frac{h(-4x -2h -1)}{h} = -4x -2h -1$

Answer:

$-4x - 2h - 1$