Let f(x)=casesx^3+8, & x

Question

Let $f(x)=\begin{cases}x^3+8, & x<0\ x^2-4, & x\geq0\end{cases}$ and $g(x)=\begin{cases}(x-8)^{1/3}, & x<0\ (x+4)^{1/2}, & x\geq0\end{cases}$.

Then the number of points where the function $g\circ f$ is discontinuous is \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$(g\circ f)(x)=g(f(x))$, so we must determine, for each $x$, the sign of $f(x)$ to know which branch of $g$ applies.

For $x<0$, $f(x)=x^3+8$. This is negative when $x^3<-8$, i.e. $x<-2$. So for $-2\le x<0$, $f(x)\geq0$, and for $x<-2$, $f(x)<0$.

For $x\geq0$, $f(x)=x^2-4$. This is negative when $0\le x<2$, and non-negative for $x\ge2$.

Piecing together $(g\circ f)(x)$:

- $x<-2$: $f(x)<0\Rightarrow g(f(x))=((x^3+8)-8)^{1/3}=x$
- $-2\le x<0$: $f(x)\geq0\Rightarrow g(f(x))=\sqrt{(x^3+8)+4}=\sqrt{x^3+12}$
- $0\le x<2$: $f(x)<0\Rightarrow g(f(x))=((x^2-4)-8)^{1/3}=(x^2-12)^{1/3}$
- $x\geq2$: $f(x)\geq0\Rightarrow g(f(x))=\sqrt{(x^2-4)+4}=\sqrt{x^2}=x$

Checking the joins $x=-2,0,2$:

- At $x=-2$: $\lim_{x	o-2^-}x=-2$, $\lim_{x	o-2^+}\sqrt{x^3+12}=\sqrt{4}=2$. Unequal $\Rightarrow$ discontinuous.
- At $x=0$: $\lim_{x	o0^-}\sqrt{x^3+12}=\sqrt{12}$, $\lim_{x	o0^+}(x^2-12)^{1/3}=(-12)^{1/3}$. Unequal $\Rightarrow$ discontinuous.
- At $x=2$: $\lim_{x	o2^-}(x^2-12)^{1/3}=(-8)^{1/3}=-2$, $\lim_{x	o2^+}x=2$. Unequal $\Rightarrow$ discontinuous.

So $g\circ f$ is discontinuous at exactly $x=-2,0,2$: 3 points.

Solutions · 1