Answer
$f+g=4x^{2}+2x-1$
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$f-g=-2x^{2}+2x+1$
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$fg=3x^{4}+6x^{3}-x^{2}-2x$
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$f/g=\dfrac{x^{2}+2x}{3x^{2}-1}$
$D:\{x|x\ne\dfrac{\sqrt{3}}{3},-\dfrac{\sqrt{3}}{3}\}$ or $D:(-\infty,-\dfrac{\sqrt{3}}{3})\cup(-\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3})\cup(\dfrac{\sqrt{3}}{3},\infty)$
Work Step by Step
$f(x)=x^{2}+2x,$ $g(x)=3x^{2}-1$
$f+g$
$f(x)+g(x)=x^{2}+2x+3x^{2}-1=4x^{2}+2x-1$
The domain of this function includes all real numbers. It can be expressed as follows:
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$f-g$
$f(x)-g(x)=x^{2}+2x-(3x^{2}-1)=x^{2}+2x-3x^{2}+1=...$
$...=-2x^{2}+2x+1$
The domain of this function includes all real numbers. It can be expressed as follows:
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$fg$
$f(x)\cdot g(x)=(x^{2}+2x)(3x^{2}-1)=3x^{4}-x^{2}+6x^{3}-2x=...$
$...=3x^{4}+6x^{3}-x^{2}-2x$
The domain of this function includes all real numbers. It can be expressed as follows:
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$f/g$
$\dfrac{f(x)}{g(x)}=\dfrac{x^{2}+2x}{3x^{2}-1}$
This function is undefined for the values of $x$ that make the denominator equal to $0$. To find these values, let's solve the following equation:
$3x^{2}-1=0$
$3x^{2}=1$
$x^{2}=\dfrac{1}{3}$
$\sqrt{x^{2}}=\sqrt{\dfrac{1}{3}}$
$x=\pm\dfrac{1}{\sqrt{3}}=\pm\dfrac{\sqrt{3}}{3}$
Now, we know the values for which this function is undefined. Therefore, this values are not included in its domain, which can be expressed as follows:
$D:\{x|x\ne\dfrac{\sqrt{3}}{3},-\dfrac{\sqrt{3}}{3}\}$ or $D:(-\infty,-\dfrac{\sqrt{3}}{3})\cup(-\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3})\cup(\dfrac{\sqrt{3}}{3},\infty)$
