Determine the domain of the following functions.
f\left(x\right)=\left(2x^2-3\right)^{-1/4}
We have:
f\left(x\right)=\left(2x^2-3\right)^{-1/4} = \dfrac{1}{\sqrt[4]{2x^2-3}}
Since the index of the radical is even, the expression inside the radical must be greater than zero. Notice that it cannot be zero as it makes the denominator zero. The domain of f is satisfying the following:
2x^2-3 \gt 0 \\x^2 \gt \dfrac{3}{2}
We have either:
\begin{cases} x \lt -\sqrt{ \dfrac{3}{2}} \cr \cr \text{or} \cr \cr x \gt \sqrt{ \dfrac{3}{2}} \end{cases}
The domain of the function is \left( -\infty,-\sqrt{ \dfrac{3}{2}} \right) \cup \left( \sqrt{ \dfrac{3}{2}}, \infty \right) .
f\left(x\right)=\sqrt{2x-4}+4\sqrt{x+3}
When the index of the radical is even, the expression inside the radical must be greater than or equal to zero. The domain of f is satisfying the following:
2x-4\geqslant0
And:
x+3\geqslant0
2x-4\geqslant0 is equivalent to:
x\geqslant2
x+3\geqslant0 is equivalent to:
x\geqslant-3
Intersecting the two solutions, we get:
x\geqslant2
The domain of the function is \left[2, \infty\right).
f\left(x\right)=\sqrt[3]{\frac{x^{2}+2x-3}{x+5}}
When the index of the radical is odd, there are no restrictions on the domain.
However, since there is a fraction, the denominator cannot equal zero.
Thus the domain of f is satisfying the following:
x+5\neq0
Which is equivalent to:
x\neq-5
The domain of the function is \left( -\infty,-5 \right)\cup\left( -5,\infty \right).
f\left(x\right)=\sqrt[3]{x}+4\sqrt{x-2}
When the index of the radical is even, the expression inside the radical must be greater than or equal to zero.
When the index of the radical is odd, there are no restrictions on the domain.
Thus the domain of f is satisfying the following:
x-2\geqslant0
which is equivalent to:
x\geqslant2
The domain of the function is \left[ 2,\infty \right).
f\left(x\right)=\left(x^{2}+4x-5\right)^{\frac{1}{2}}
We have:
f\left(x\right)=\left(x^{2}+4x-5\right)^{\frac{1}{2}}=\sqrt{x^{2}+4x-5}
When the index of the radical is even, the expression inside the radical must be greater than or equal to zero.
Thus the domain of f is satisfying the following:
x^{2}+4x-5\geqslant0
which is equivalent to:
\left(x-1\right)\left(x+5\right)\geqslant0
The solutions of the equation attached to the inequality are:
x=1 and x=-5
Since the coefficient of x^{2} is positive, the solution is outside -5 and 1.
The domain of the function is \left( -\infty,-5 \left]\cup\right[ 1,\infty \right).
f\left(x\right)=\left(x^{2}-9\right)^{-\frac{1}{3}}
We have:
f\left(x\right)=\left(x^{2}-9\right)^{-\frac{1}{3}}=\dfrac{1}{\sqrt[3]{x^{2}-9}}
When the index of the radical is odd, there are no restrictions on the domain.
However, since there is a fraction, the denominator cannot equal zero.
So the domain of f is satisfying the following:
x^{2}-9\neq0
which is equivalent to:
x^{2}\neq9
x\neq3 or x\neq-3
The domain of the function is \left( -\infty,-3 \right)\cup\left(-3{,}3\right)\cup\left( 3,\infty \right).
f\left(x\right)=\sqrt{\dfrac{x+1}{x-5}}
When the index of the radical is even, the expression inside the radical must be greater than or equal to zero.
Since there is a fraction, the denominator cannot equal zero:
x\neq5
So the domain of f is satisfying the following:
\dfrac{x+1}{x-5}\geqslant0
which is equivalent to:
x \leq -1 or x \gt 5
The domain of the function is \left(-\infty,-1 \right]\cup\left( 5,\infty \right).