Answer
a. $y'=-2x$
b. See graph.
c. $f'(x)\gt0$ when $x\in(-\infty,0)$, $f'(0)=0$, and $f'(x)\lt0$ when $x\in(0, \infty)$
d. $x\in(-\infty,0)$, $x\in(0, \infty)$.
Work Step by Step
a. Given $y=f(x)=-x^2$, we have $y'=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{-(x+h)^2+x^2}{h}=\lim_{h\to0}\frac{-2xh-h^2}{h}=\lim_{h\to0}(-2x-h)=-2x$
b. See graph for the function and its derivative.
c. As can be identified from the graph, $f'(x)\gt0$ when $x\in(-\infty,0)$, $f'(0)=0$, and $f'(x)\lt0$ when $x\in(0, \infty)$
d. Over the intervals of $x\in(-\infty,0)$, the function $y=f(x)$ increases as x increases.
Over the intervals of $x\in(0, \infty)$, the function $y=f(x)$ decreases as x increases.
When the function increases when x increases, the derivative will be positive as shown in part-c;
when the function decreases when x increases, the derivative will be negative as shown in part-c.
