Prove that $(1+a)^x>1+ax$ when $x>1$ and $0Prove that $(1+a)^x>1+ax$ when $x>1$ and $0<a<1$ and $(1+a)^x<1+ax$ when $0<x<1$ and $0<a<1$
I was trying to do it the usual way which is to consider the function $h(x)=(1+a)^x-1-ax$. Note that when $x=1$, $h(x)=0$; taking the derivative $h´(x)=(1+a)^xln(1+a)-a$, and I´d like to prove that $h´(x)>0$ when $x>1$, and this would imply that $h(x)$ is an increasing function which would also imply that $h(x)>0$
But that´s my problem how can I guarantee/prove that $h´(x)>0$ when $x>1$ ? I would really appreciate your help
 A: Hint: Study it as a function of $a$, not $x$.
A: You can proof this by taking the Taylor series of the left hand side with respect to $a$ at $a=0$,
$$
(1+a)^x=1+xa+\frac{x}{2}(x-1)a^2+O(a^3).
$$
A: Use the mean value theorem. Let $x>0$ and define for $t>-1$
$$f(t):=(1+t)^x\;.$$
Given $a>0$, consider $f(a)-f(0)$. By the MVT, there exists $b$ between $0$ and $a$ such that
$$
(1+a)^x-1=f(a)-f(0)\stackrel{MVT}=(a-0)f'(b) = ax(1+b)^{x-1}.\tag{*}
$$
Note that $1+b>1$. So if $x>1$ then $(1+b)^{x-1}>1$; if $0<x<1$ then $(1+b)^{x-1}<1$. Since every factor on the RHS of (*) is positive, we're done!
A: Hint: To prove the special case, where $r \in \mathbb{R}$ such that $r \geq -1$ and $N$ is a positive integer, use
$$\sum_{k = 0}^{N - 1} r^k = \frac{1- r^{N}}{1 - r} = \frac{r^{N} - 1}{r - 1}$$
If $R = 1 + r$, then $(1 + r)^N = 1 + (R^N - 1)$. Now study
$$\sum_{k = 0}^{N - 1} R^k = \frac{R^{N} - 1}{R - 1}.$$
However, to prove your statement use the Taylor series for $(1 + a)^x$.
A: $(1+a)^x=1+ax$ if $x=0$ and if $x=1$. $(1+a)^x$ is strictly convex if $a>0$, at the same time $1+ax$ is linear for all $x$. As a result, outside of the interval $(0,1)$ $(1+a)^x>1+ax.$ For the same reason $(1+a)^x<1+ax$ within $(0,1)$.

$(1+a)^x=e^{x\ln(1+a)}$; $\frac{d}{dx}(1+a)^x=\ln(1+a)e^{x\ln(1+a)}$; $\frac{d^2}{dx^2}(1+a)^x=\ln^2(1+a) e^{x\ln(1+a)}>0.$
