
How do you describe the intervals(s) on which the below function is continuous, using interval notation?\[f(x)=x\sqrt(x+6)\]
Answer
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Hint: 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:
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.
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 \[\left[ -6,\infty \right)\].
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