Question

given the function $f(x) = 3 - 2x^2$, evaluate $dfrac{f(x + h) - f(x)}{h}$, $h \
eq 0$. select one: \LXI0 a. $-4x - 2h^2$ \LXI1 b. $4x + 2h$ \LXI2 c. $-4x + 2h$ \LXI3 d. $-4x - 2h$

Explanation:

Step1: Compute $f(x+h)$

$f(x+h) = 3 - 2(x+h)^2 = 3 - 2(x^2 + 2xh + h^2) = 3 - 2x^2 - 4xh - 2h^2$

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

$f(x+h)-f(x) = (3 - 2x^2 - 4xh - 2h^2) - (3 - 2x^2) = -4xh - 2h^2$

Step3: Divide by $h$ ($h

eq0$)
$\frac{f(x+h)-f(x)}{h} = \frac{-4xh - 2h^2}{h} = -4x - 2h$

Answer:

D. $-4x - 2h$