For $x\in \mathbb{R}$ I have two continuous and differentiable real-valued functions $f(x)$, $g(x)$, which are single-peaked and maximized at $x^f$ and $x^g$ respectively, with $x^g<x^f$. Let $\alpha>0$ and suppose that I want to maximize a function $v(x)$, such that:
$$v(x)=f(x)+\alpha g(x)$$
I conjecture that the value of $x$ that maximizes $v(x)$, call it $x^v$, is such that:
$$x^v=\frac{x^f}{\alpha}+x^g.$$
I would like to know whether you think my conjecture is correct, and under which conditions it makes sense. I think it could possibly just come as an application of the mean value theorem, is that correct? I do not want to waste time proving something false.
Edit due to answer below
Indeed, it seems like a better conjecture would be
$$x^v=\frac{x^f+\alpha x^g}{1+\alpha}.$$
For example, the functions $f(x)=-(x-x^f)^2$ and $g(x)=-(x-x^g)^2$ clearly yield the above formula. Notice that $x^v(\alpha=0)=x^f$, $\lim_{\alpha\to \infty }x^v(\alpha)=x^g$, and $x^v\in [x^f,x^g)$.
Your conjecture does not seem correct, take for example $f(x)=-x^2$, $g(x)=-(x+1)^2$, which respectively peak at $x^f=0$ and $x^g=-1$. Then $f(x)+\alpha g(x)=-(1+\alpha)x^2-2\alpha x-\alpha$ has a maximum at $x=-\frac{\alpha}{1+\alpha}$, which, unless $\alpha=-\frac{1}{2}$, is not equal to $1=\frac{x^f}{\alpha}+x^g$.
Edit
I think the second conjecture $x^v=\frac{x^f+\alpha x^g}{1+\alpha}$ is not correct either. It does work for square functions, but does not work for functions of the form $(x-x_0)^{2n}$ with $n\geq 2$. Take for example $f(x)=-x^{2n}$, $g(x)=-(x+1)^{2n}$, then $v(x)=-x^{2n}-\alpha(x+1)^{2n}$ and $v'(x)=-2n\left(x^{2n-1}-\alpha(x+1)^{2n-1}\right)$.
But $v'\left(\frac{x^f+\alpha x^g}{1+\alpha}\right)=v'\left(-\frac{\alpha}{1+\alpha}\right)=\frac{2n}{(1+\alpha)^{2n-1}}\left(\alpha^{2n-1}+\alpha\right)\neq 0$, so $v$ cannot have a maximum at $\frac{x^f+\alpha x^g}{1+\alpha}$.
Reasoning with functions of the form $f(x)=-(x-x^f)^{2n}$ and $g(x)=-(x-x^g)^{2n}$, the maximum of $v(x)=f(x)+\alpha g(x)$ is attained at $x=\frac{x^f+\alpha^{1/2n-1}x^g}{1+\alpha^{1/2n-1}}$, which gives back your formula when $n=1$. I doubt there exists a formula for more general functions.