Determining if a function is strictly increasing for all natural numbers on a specific interval I am having some difficulties with tackling the following problem:
Show that $f(x)=x^n$ (where n ∈ $\mathbb{N}$) is strictly increasing on the interval $[0, ∞)$.
I am familiar with the following definition: Strictly increasing means if $x_1 < x_2$ then $f(x_1) < f(x_2)$. So, I started with this definition. However, I am not sure if this is supposed to be proved using the mathematical induction method. I tried to split it into two cases: an even and an odd exponent, but this didn't really lead me anywhere, unfortunately.
 A: If you know how to use induction, then proceed like this: $f(x)=x^1=x$ is clearly strictly increasing, because if $0 \leq x<y$, then $f(x)=x<y=f(y)$. Now assume the result is true for $n=1,2,\ldots,k$. We need show this is true for $n=k+1$. 
Suppose $0 \leq x<y$. Let $f(x)=x^{k+1}$ and $g(x)=x^k$. By assumption, we know that $g(x)$ is increasing so that $g(x)= x^k<y^k=g(y)$. But
$$
f(x)= x^{k+1}= x \cdot x^k= x \cdot g(x) \leq y \cdot g(x) < y \cdot g(y)= y \cdot y^k= y^{k+1}= f(y)
$$
But then $f(x)$ is increasing.
A: hint
Let $h$ and $g$ be two stricly increasing functions from $(0,+\infty)$ to $(0,+\infty)$.
then, for $x,y>0$
$$x<y \implies h(x)<h(y) $$
$$\implies h(x)g(x)<h(y)g(x)<h(y)g(y)$$
thus
the product $ hg$ is stricly increasing at $(0,+\infty)$.
Use the fact that $f(x)=x.x.x...x$
and $f(0)=0$ to conclude that $f$ is stricly increasing at $[0,+\infty)$.
A: A strictly increasing function $f$ on $(0,\infty)$ satisfies that for $x_1<x_2$, $f(x_1)<f(x_2)$.
Let $x_1=x$ and $x_2=x+\varepsilon$.  Setting $\varepsilon>0$ satisfies the inequality $x_1<x_2$.  We wish to show that $x^n<(x+\varepsilon)^n$ for all $n\in\Bbb N$.
By the binomial theorem, we have 
\begin{align}
(x+\varepsilon)^n&=\sum_{r=0}^n\binom{n}{r} x^{n-r}\varepsilon^r\\
&=\binom{n}0x^n+\binom{n}1x^{n-1}\varepsilon+\cdots+\binom{n}{n-1}x\varepsilon^{n-1}+\binom{n}{n}\varepsilon^n\\
&=x^n+\mathcal{P}(x)>x^n
\end{align}
Since we have $x>0$ and $\varepsilon>0$, it is easy to see that $\mathcal{P}(x)>0$ since the combination, the power of $x$ and the power of $\varepsilon$ are all positive.
A: We prove by induction.
Base case ($n=1$)
$f_1(x)=x$, this is the identity function, so for $x_1<x_2$ this property hold trivially.
Induction Hypothesis
Let us assume that $f_{n-1}(x)=x^{n-1}$ is strictly increasing on $[0,\infty)$.
Now we want to prove for $f_n(x)=x^n$, but we note that $f_n(x)=f_{n-1}(x)f_1(x)$, by the IH and base case these are two strictly increasing functions, so their product is also strictly increasing (Why?)
And we're done
A: 1) $x>0$;
$x_1=x$, $x_2= x+h$, $h >0$;
$x_2^n = x^n(1+h/x)^n \ge$
$x^n(1+n(h/x))$;
$x_2^n-x_1^n=$
$x^n(1+h/x)^n-x^n \ge$
$x^n(1+n(h/x))-x^n=$
$nh x^{n-1} >0$.
2) $x=0$; $h>0$;
$x_2^n-x_1^n=h^n >0$.
A: I just taught this today.
Let $0\leq x_1<x_2$.
Notice that
$x_2^n-x_1^n=(x_2-x_1)(\sum_{i=0}^{n-1}x_1^i x_2^{n-1-i})$ is positive since both factor are. Hence $x_2^n>x_1^n$ and the function is increasing.
