Answer
a. {$x$|$x$ is a real number}
b. $(-∞,∞)$
Work Step by Step
We are given the function $f(x)=\frac{9x^{2}-8x+3}{4x^{2}+1}$. The domain of the function will be all values of $x$ such that the denominator does not equal 0.
Therefore, we can set the denominator equal to 0. We will exclude from the domain all values of $x$ that make the denominator equal 0.
$4x^{2}+1=0$
Subtract 1 from both sides.
$4x^{2}=-1$
Divide by 4.
$x^{2}=-\frac{1}{4}$
There is no real number value for $x$ such that $x^{2}=-\frac{1}{4}$, so we know that the domain includes all real numbers.
Therefore, the domain is {$x$|$x$ is a real number} in set-builder notation and $(-∞,∞)$ in interval notation.