Question

given the function $f(x) = -3x^2$
find the difference quotient $\frac{f(x + h) - f(x)}{h}$
find $f(x)$ by determining $limlimits_{h\to 0} \frac{f(x + h) - f(x)}{h}$.
find $f(-2)$
find $f(0)$
find $f(1)$
graph the function $f(x) = -3x^2$ and draw the tangent lines to the graph at points whose x-coordinates are -2, 0, and 1. (the slopes of these lines should match the derivative values you calculated above.)

Explanation:

Part 1: Find the difference quotient $\frac{f(x + h) - f(x)}{h}$

Step1: Compute $f(x + h)$

Given $f(x) = -3x^2$, substitute $x$ with $x + h$:
$f(x + h) = -3(x + h)^2 = -3(x^2 + 2xh + h^2) = -3x^2 - 6xh - 3h^2$

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

Subtract $f(x)$ from $f(x + h)$:
$f(x + h) - f(x) = (-3x^2 - 6xh - 3h^2) - (-3x^2) = -3x^2 - 6xh - 3h^2 + 3x^2 = -6xh - 3h^2$

Step3: Divide by $h$

Divide the result by $h$ (assuming $h
eq 0$):
$\frac{f(x + h) - f(x)}{h} = \frac{-6xh - 3h^2}{h} = \frac{h(-6x - 3h)}{h} = -6x - 3h$

Answer:

(Difference Quotient): $-6x - 3h$

Part 2: Find $f'(x)$ by $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step1: Use the difference quotient result

From part 1, $\frac{f(x + h) - f(x)}{h} = -6x - 3h$

Step2: Take the limit as $h \to 0$

$\lim_{h \to 0} (-6x - 3h) = -6x - 3(0) = -6x$