Question

How do you describe the intervals(s) on which the below function is continuous, using interval notation?

$f(x)=x\sqrt(x+6)$

Answer 1

For a function which is a product of two or more functions to be continuous. For example, $f(x)=g(x)h(x)$, for $f(x)$ to be continuous $g(x)$ and $h(x)$ both need to be continuous. So, we can find the range on which $f(x)$ is continuous by finding separately the range on which $g(x)$ and $h(x)$ both are defined and then taking the intersection of the ranges.

Complete step by step answer:

The given function is $f(x)=x\sqrt(x+6)\Rightarrow x\sqrt{x+6}$. It is of the form of $f(x)=g(x)h(x)$. We know that to find the range on which $f(x)$ is continuous we need to separately find the range on which $g(x)$ and $h(x)$ both are defined, and then we take the intersection of the ranges. We will do the same for this function.

Here, $g(x)=x$ and $h(x)=\sqrt{x+6}$,

For the function $g(x)=x$, as it is a linear function, it is defined for $\left( -\infty ,\infty \right)$.

For the function $h(x)=\sqrt{x+6}$, it is of the form $\sqrt{a}$. We know that for $\sqrt{a}$ to be defined, the only condition required is $a$ must be a non-negative quantity. Hence for $h(x)=\sqrt{x+6}$, to be defined $x+6\ge 0$. Subtracting 6 from both sides of this equation, we get

$\Rightarrow x+6-6\ge 0-6$

$\Rightarrow x\ge -6$

So for $h(x)=\sqrt{x+6}$ to be defined, the range is $\left[ -6,\infty \right)$.

We have range for both functions, we need to take their intersection

$\Rightarrow \left( -\infty ,\infty \right)\bigcap \left[ -6,\infty \right)$

$\Rightarrow \left[ -6,\infty \right)$

Hence the range on which $f(x)=x\sqrt(x+6)$ is continuous is $\left[ -6,\infty \right)$.

Note: We can do the same thing for addition, subtraction, or division of the function. That is for functions of the form $f(x)=g(x)+h(x)$, $f(x)=g(x)-h(x)$, $f(x)=\dfrac{g(x)}{h(x)}$. Here for the function of the form $f(x)=\dfrac{g(x)}{h(x)}$, the values for which $h(x)=0$ should be excluded from the range.