Let f(x)=x^3+8; & xThen the number of points, where the function g f is discontinuous, is

Question

Let $f(x)={x}^{3}+8; \& xThen the number of points, where the function $g\circ f$ is discontinuous, is $

Accepted Answer

We need to find the points where the composite function $g \circ f$ is discontinuous.

That means we first find

$$ (g \circ f)(x)=g(f(x)). $$

So we must see, for each value of $x$, which branch of $g$ will be used.

1. Given functions
$$ f(x)= \begin{cases} x^3+8, & xand

\end{cases}$$ g(x)=  (x-8)\^{1/3}, \& xNow,

$$ (g\circ f)(x)=g(f(x)). $$

The definition of $g$ depends on whether its input is negative or non-negative. So we must check the sign of $f(x)$.

2. Find where $f(x)
Case 1: $x
Then

$f(x)=x^3+8.$

We check when this is negative:

$$ x^3+8So for $x
if $x
if $-2\le x

Case 2: $x\ge 0$
Then

$$ f(x)=x\^2-4. $We check when this is negative:$ x\^2-4But here $x\ge 0$, so this becomes

$$ 0\le xThus for $x\ge 0$:

if $0\le x
if $x\ge 2$, then $f(x)\ge 0$.

3. Write $(g\circ f)(x)$ piecewise
Now use the correct branch of $g$.

(i) When $x
Here $f(x)$$ g(y)=(y-8)\^{1/3}. $Thus$ (g\circ f)(x)=\big((x\^3+8)-8\big)\^{1/3}=(x\^3)\^{1/3}=x. $So for $x$ (g\circ f)(x)=x. $$

(ii) When $-2\le x
Here $f(x)\ge 0$, so use

$$ g(y)=\sqrt{y+4}. $Thus$ (g\circ f)(x)=\sqrt{(x\^3+8)+4}=\sqrt{x\^3+12}. $$

So for $-2\le x$$ (g\circ f)(x)=\sqrt{x\^3+12}. $$

(iii) When $0\le x
Here $f(x)$$ g(y)=(y-8)\^{1/3}. $Thus$ (g\circ f)(x)=\big((x\^2-4)-8\big)\^{1/3}=(x\^2-12)\^{1/3}. $$

So for $0\le x$$ (g\circ f)(x)=(x\^2-12)\^{1/3}. $$

(iv) When $x\ge 2$
Here $f(x)\ge 0$, so use

$$ g(y)=\sqrt{y+4}. $Thus$ (g\circ f)(x)=\sqrt{(x\^2-4)+4}=\sqrt{x\^2}=x $$

since $x\ge 2$.

So for $x\ge 2$,

$$ (g\circ f)(x)=x. $$

4. Check continuity at possible boundary points
Each expression is continuous in its own interval. So we only check the joining points:

$$ x=-2,\quad x=0,\quad x=2. $At $x=-2$
Left-hand limit:

for $x$ \lim\_{x\to -2\^-}(g\circ f)(x)=-2. $$

Right-hand limit:

for $-2\le x$$ \lim\_{x\to -2\^+}(g\circ f)(x)=\sqrt{(-2)\^3+12}=\sqrt{4}=2. $Since$ -2\ne 2, $$

the function is discontinuous at $x=-2$.

At $x=0$
Left-hand limit:

$$ \lim\_{x\to 0\^-}(g\circ f)(x)=\sqrt{0\^3+12}=\sqrt{12}. $Right-hand limit:$ \lim\_{x\to 0\^+}(g\circ f)(x)=(0\^2-12)\^{1/3}=(-12)\^{1/3}. $$

These are not equal, so discontinuous at $x=0$.

At $x=2$
Left-hand limit:

$$ \lim\_{x\to 2\^-}(g\circ f)(x)=(2\^2-12)\^{1/3}=(-8)\^{1/3}=-2. $Right-hand limit:$ \lim\_{x\to 2\^+}(g\circ f)(x)=2. $$

These are not equal, so discontinuous at $x=2$.

5. Conclusion
The function $g\circ f$ is discontinuous at

$$ x=-2,\quad x=0,\quad x=2. $$

Hence, the number of points of discontinuity is

$$

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