Show that $f(x)=x^r+r^x$ is increasing

Question

I know that the function $f(x)=x^r+r^x$ is increasing where $0<r<1$ and $x\geq 2$ by the graph. How to show it? I couldn't prove it .

Answer 1

We have to show that $rx^{r-1}+r^x\log(r) > 0$ for $r\in (0,1)$ and $x\ge 2$. This inequality is equivalent to $\tfrac{x^{r-1}}{r^{x-2}(-r\log(r))} > 1$, i.e., $$ \frac{1}{-r\log(r)}e^{(r-1)\log(x)-(x-2)\log(r)} > 1. $$ The LHS is positive, so it is valid to apply the logarithm: $$ -\log(-r\log(r)) + (r-1)\log(x)-(x-2)\log(r) > 0. $$ The first summand is independent of $x$ and is easily seen to be larger than $1$ for $r\in (0,1)$. Hence, it suffices to show that $h_r(x)\ge -1$, where $$ h_r(x) := (r-1)\log(x)-(x-2)\log(r). $$ Since $r-1 < 0$ and $\log(x)\le x-1$, we have $$ h_r(x)\ge (r-1)(x-1)-(x-2)\log(r) = (x-2)(r-1-\log(r)) + r-1. $$ Now, have a look at $g(r) = r-1-\log(r)$. We have $g(0+) = +\infty$, $g(1-) = 0$, and $g'(r) = 1-\tfrac 1 r < 0$. Thus $g(r) > 0$ for $r\in (0,1)$ and $h_r(x)\ge r-1 > -1$ follows for $x\ge 2$.

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With this analysis we can say more: as shown above, the function $f_r(x) = x^r+r^x$ increases at $x$ if and only if $-\log(-r\log(r)) + h_r(x) > 0$. As $h_r$ is convex and its minimum is at $x=m(r):=\tfrac{r-1}{\log(r)}$, the function $f_r$ increases on $[0,\infty)$ if and only if $G(r) := -\log(-r\log(r)) + h_r(m(r)) > 0$. A plot shows that $G$ is an increasing function on $(0,1)$ and that its zero is somewhere close to $0.39$. In particular, $f_r$ is increasing on $[0,\infty)$ for all $r\ge 0.4$.

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Moreover, $$ -\log(-r\log(r)) + h_r(x)\ge (r-1-\log(-r\log(r))) + (x-2)(r-1-\log(r)). $$ Hence, $-\log(-r\log(r)) + h_r(x) > 0$ if $$ 2-x < \frac{r-1-\log(-r\log(r))}{r-1-\log(r)}. $$ A plot shows that the function on the RHS is bounded below by $0.38$. Hence, $f_r$ is even increasing on $[1.62,\infty)$ for all $r\in (0,1)$.

Answer 2

Hint: You can compute the first derivative $$f'(x)=rx^{r-1}+r^x\ln(r)$$ but the summand $r^x\ln(r)$ is negative for $$0<r<1$$

Answer 3

We need $$f'(x)=rx^{r-1}+\ln r\cdot r^x > 0.$$ This is the same as $$ rx^{r-1}>\ln(1/r)\cdot r^x. $$ Rewrite the RHS: $$\ln(1/r)\left(1-x\ln(1/r) - \frac12x^2\ln^2(1/r) + O(x^3)\right).$$ Can you continue?