## University Calculus: Early Transcendentals (3rd Edition)

$f$ is both continuous and differentiable at $x=0$.
a) The function is graphed below. b) To test the continuity of the function at $x=0$, we examine $\lim_{x\to0^+}f(x)$ and $\lim_{x\to0^-}f(x)$. - For $x\to0^+$, since $x\gt0$, we use the function $f(x)=x^2$: $$\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x^2=0^2=0$$ - For $x\to0^-$, since $x\lt0$, we use the function $f(x)=-x^2$: $$\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-x^2)=(-0)^2=0$$ As $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$, we conclude that $\lim_{x\to0}f(x)$ exist and hence $f$ is continuous at $x=0$. c) To test the differentiability of the function at $x=0$, we also examine the left-hand and right-hand derivative function $\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h}$. We have $$\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\frac{f(0)-f(0)}{h}=0$$ $$\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\frac{f(0)-f(0)}{h}=0$$ As $\lim_{h\to0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0^+}\frac{f(0+h)-f(0)}{h}$, we conclude that $f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}$ exist and hence $f$ is differentiable at $x=0$.