Solution of $3^x+x^2=1$ using calculus I'm trying to find out the solutions of $3^x+x^2=1$ using calculus. If we double differentiate both sides and then take the logarithm, we get the value of $x$. 
After double differentiating, the equation becomes :
$$3^x \ln^2 3 + 2=0$$ and we can take the logarithm of $3^x$ and do that way. But is that a correct way? I doubt that. If not, why? 
Also, how to calculate the values? 
 A: Note that $x=0$ is a solution to the equation $3^x+x^2=1$. 
Consider the function $f(x)=3^x+x^2-1$. Note that this function is positive for $x>0$, and it is positive for $x\leq -1$. Thus, the roots of this equation must belong to the set $(-1,0]$. The second derivative of this function is $f''(x)=3^x\ln(3)^2+2>0$, so we know that the first derivative $f'(x)=3^x\ln(3)+2x$ is strictly increasing and can therefore only have one root. Since $f'(-1)<0$ and $f'(0)>0$, by the intermediate value theorem there is a $c\in(-1,0]$ such that $f'(c)=0$. Thus, on each interval $(-\infty,c)$ and $[c,\infty)$ the function $f(x)$ can have at most one root.
We have already found the root on the interval $[c,\infty)$, so what remains is to find the root on $[-1,c)$. Since we know that $f(x)$ is strictly decreasing on that interval, and $f(-1)>0$, you could implement Newton's method with a starting guess of $x_0=-1$ in order to approximate the root.
Can we instead find a closed form expression for this root? No. This function does not permit us to solve for $x$ analytically; the best we will ever do is write down the solution in terms of special functions, which will also necessitate the use of approximations at the end of the day.
