Determine the domain of the functions given by the following expressions.
f\left(x\right)=\dfrac{2x^2+3x}{3x^2-2}
The domain of this equation is the set of all real numbers except those which make the denominator zero.
We have:
3x^2 - 2 = 0 \leftrightarrow x^2= \dfrac{2}{3} \leftrightarrow x =\pm\sqrt{\dfrac{2}{3}}
Therefore the domain is:
\mathbb{R}- \left\{ \sqrt{\dfrac{2}{3}},-\sqrt{\dfrac{2}{3}} \right\}
f\left(x\right)=\dfrac{4x+1}{x^2+2}
The domain of this equation is the set of all real numbers except those which make the denominator zero. Observe that for every real number x we have:
x^2+2 \gt 0
Therefore, the denominator is always non-zero and the domain of f equals:
\mathbb{R}
f\left(x\right)=\dfrac{2-x^2}{4x+x^3}
The domain of this equation is the set of all real numbers except those which make the denominator zero.
We have:
4x+x^3 = 0
x\left(4+x^2\right)=0
\begin{cases}x=0 \cr \cr or \cr \cr 4+x^2=0 \end{cases}
The second equation has no real solution. Therefore, x=0 is the only root of the denominator.
The domain is:
\mathbb{R} -\{0\}
f\left(x\right)=\dfrac{x^3-7x^2+1}{4x^2-25}
The domain of this equation is the set of all real numbers except those which make the denominator zero.
We have:
4x^2 - 25 = 0
\Leftrightarrow x^2= \dfrac{25}{4}
\Leftrightarrow x =\pm {\dfrac{5}{2}}
The domain is:
\mathbb{R}- \left\{ {-\dfrac{5}{2}},{\dfrac{5}{2}} \right\}
f\left(x\right)=\dfrac{x^2-1}{x^2-5x+6}
The domain of this equation is the set of all real numbers except those which make the denominator zero.
We have:
x^2 - 5x + 6 = 0
We can factor this equation as follows:
x^2 - 5x + 6 = \left(x-3\right)\left(x-2\right) = 0
The equation contains two roots, namely 2 and 3.
Therefore, the domain is:
\mathbb{R} - \{ 2{,}3}\
f\left(x\right)=\dfrac{x}{x^2-2x+1}
The domain of this equation is the set of all real numbers except those which make the denominator zero.
We have:
x^2 - 2x+1 = 0
We can factor it as follows:
x^2-2x+1=\left(x-1\right)^2
We can write:
\left(x-1\right)^2=0
\Leftrightarrow x=1
Therefore, the domain is:
\mathbb{R}- \{1\}
f\left(x\right)=\dfrac{2x-1}{x^2+x^3}
The domain of this equation is the set of all real numbers except those which make the denominator zero.
We have:
x^2+x^3=0
x^2\left(1+x\right) = 0
\begin{cases}x^2=0 \cr \cr or \cr \cr x+1=0 \end{cases}
From the first equation we have x=0, and from the second we have x=-1.
Therefore, the domain is:
\mathbb{R}- \{-1{,}0\}