Diffrenece of exponential functions

Question

Prove that the function $f: \Bbb R \to \Bbb R$, $f(x)=2016^x-2015^x+x$ is strictly increasing. I tried to find the derivative, but it didn't help me.

Answer 1

We have $$f'(x)=2016^x\ln 2016-2015^x\ln 2015 +1$$ $$f''(x)=2016^x\ln^2 2016-2015^x\ln^22015$$ Now $$\begin{align}f''(x)\ge 0&\iff 2016^x\ln^2 2016-2015^x\ln^22015\ge 0\\&\iff 2016^x\ln^2 2016\ge 2015^x\ln^22015\\&\iff \left(\frac{2016}{2015}\right)^x\ge \left(\frac{\ln2015}{\ln2016}\right)^2\\&\iff x\ge \log_{2016/2015}\left(\frac{\ln2015}{\ln2016}\right)^2=:\alpha\end{align}$$

So, we know that $f'(x)$ is decreasing for $x\lt \alpha$ and that $f'(x)$ is increasing for $x\gt \alpha$, and so, it is sufficient for proving that $f'(x)\gt 0$ to prove that $f'(\alpha)\gt 0$.

By the way, we have $$\ln 2015-\ln 2016\gt -1\tag1$$ because $$\begin{align}(1)&\iff \ln 2016-\ln 2015\lt 1\\&\iff \ln\frac{2016}{2015}\lt \ln e\\&\iff \frac{2016}{2015}\lt e\end{align}$$ which is true.

Also, we have $$\alpha\lt 0\tag2$$ because $$\begin{align}(2)&\iff \log_{2016/2015}\left(\frac{\ln2015}{\ln2016}\right)^2\lt \log_{2016/2015}1\\&\iff \left(\frac{\ln2015}{\ln2016}\right)^2\lt 1\\&\iff \ln^2 2015\lt \ln^2 2016\\&\iff \ln 2015\lt \ln 2016\end{align}$$ which is true.

Using that $2016^\alpha=2015^\alpha\left(\frac{\ln2015}{\ln2016}\right)^2$ and $(1)(2)$, $$\begin{align}f'(\alpha)&=2016^\alpha\ln 2016-2015^\alpha\ln 2015 +1\\&=2015^\alpha\left(\frac{\ln2015}{\ln2016}\right)^2\ln2016-2015^\alpha\ln2015+1\\&=\frac{\ln2015(\ln2015-\ln2016)}{\ln2016}2015^\alpha+1\\&=\frac{\ln2015(\ln2015-\ln2016)2015^\alpha+\ln2016}{\ln2016}\\&\gt\frac{\ln2015(\ln2015-\ln2016)2015^0+\ln2016}{\ln2016}\\&\gt\frac{\ln2015(-1)+\ln2016}{\ln2016}\\&\gt 0\end{align}$$

Therefore, we have $f'(x)\gt 0$, from which we can say that $f(x)$ is strictly increasing.