Finding the values of C such that $\ln(x^2)-x^2 + C \geq 0$ I have the solution of an equation as $y=\pm \sqrt{\ln(x^2)-x^2 + C}$. I am asked to find the values of $C$ the solution is valid for. 
So I get this inequality $\ln(x^2)-x^2+C \geq 0$. How can I find the values of $C$ such that $C\geq x^2 - \ln(x^2)$.
I know that $x\neq 0$ to avoid the infinity value of the logarithm function and the root must be zero or positive.
 A: We want to have $\ln(x^2)-x^2+C \geq 0$
Write $C$ as $\ln(K)$ for $K=e^{C}$ and then combine it with $\ln(x^2)$.
$\ln(K) + \ln(x^2) = \ln(Kx^2) \geq x^2$
Since $f(x)=e^x$ is a strictly increasing function, we could apply it to both sides of the inequality to obtain:
$$Kx^2 \geq e^{x^2} \implies K \geq \frac{e^{x^2}}{x^2}$$
Since $C$ is independent of $x$ and the above inequality is valid for every $x$ in the domain of the expression $\ln(x^2)-x^2+C$, you could take limits of both sides of the inequality when $x \to 0$ and you'd get $K \geq +\infty$. Therefore, such a $C$ does not exist. 
If this proof is confusing for you, note that $x^2-\ln(x^2) \leq C$ is the same as saying that $C$ is an upper bound for $f(x)=x^2-\ln(x^2)$. But $x \to \pm\infty$ implies $f(x)\to +\infty$ and $x \to 0$ implies $f(x) \to -\infty$. Therefore, the function $f$ is unbounded. And the reason for $f(x) \to +\infty$ when $x \to \pm \infty$ that is that the rate of growth of $\ln(x^2)$ as $x$ gets large is too slow to be able to catch $x^2$. The last statement about comparing the rates of growth of two functions is intuitive and even though it's not rigorous, it's commonly used in calculus without proof.
A: Can you show that $\log(x^2) - x^2 \leq -1$?
However, as $x \to \infty$, $\log(x^2) - x^2 \to -\infty$ and you can't make it non-negative by adding a constant $C$
What is the domain of $x$? If $x$ is bounded, then such a constant $C$ might exist, depending on the maximum value of $x$
A: Let $f(x) = \ln(x^2)-x^2$. Then $f'(x)=\frac{2x}{x^2}-2x=0$ for stationary points. 
So $\frac1x=x \implies x = \pm 1$. As $f''(x)=-\frac2{x^2}-2<0$, a maximum exists.
Now $f(\pm 1) = \ln((\pm 1)^2)-(\pm 1)^2=-1$ so global maximum is $-1$. Hence a solution/solutions exist when $C \ge 1$.
