Question

Find the range values of p such that ln(3x^2 - px + 2) has domain, x is all real numbers.?

Answer 1

\(\ln(3x^2-px+2)\) will be defined for all real \(x\) if \(3x^2-px+2>0\).
\[\begin{align*}3x^2-px+2&=3\left(x^2-\frac{p}{3}x+\frac{2}{3}\right)\\\\
&=3\left(x^2-\frac{p}{3}x+\frac{p^2}{36}+\frac{2}{3}-\frac{p^2}{36}\right)\\\\
&=3\left(\left(x-\frac{p}{6}\right)^2+\frac{2}{3}-\frac{p^2}{36}\right)\\\\
&=3\left(x-\frac{p}{6}\right)^2+2-\frac{p^2}{12}
\end{align*}\]
This should help with solving for \(x\), which will tell you what values of \(p\) will work.