Question on the proof of $e^x>1+x$ for $x>0$

Question

Show that $e^x>1+x$ for $x>0$

Proof: Set $f(x)=e^x-(1+x)$. Show that $f(x)$ is always positive. We know that $f(0)=e^0-(1+0)=0$ and $f'(x)=e^x-1$. When $x$ is positive, $f'(x)$ is positive because $e^x>1$

We know that if $f'(x)>0$ on an interval, then $f(x)$ is increasing on that interval, so we can conclude that $f(x)>f(0)$ for $x>0$ and thus: $$e^x-(1+x)>0 \iff e^x>1+x$$

Question: How do we know that $f(x)$ is positive just by $e^x>1$?

Additional question, how do we prove $f'(x)$ is positive/negative over an interval $[a,b]$?

Answer 1

1. We know that $e^x$ is strictly increasing in $x$. Then, for $x>0$, the infimum value of $e^x$ is attained at $x \to 0$, which is equal to $1$. Clearly, it follows that $e^x >1$ for $x>0$.

2. The simple answer would be: If the derivative of $f(x)$ is positive/negative in a given interval, the function is increasing/decreasing in that interval. In your case, $f'(x) = e^x-1,$ which is positive for $x \in (0,1)$.

Note: for all this argument, I assume that I already know function $e^x$.

Answer 2

For $x>0$, we know $e^x > 1$, so $f'(x) > 0$. So, $f(x)$ is increasing, in particular $f(x) > f(0) = 0$, so $f(x)$ is positive.

Answer 3

How do we know that $f(x)$ is positive just by $e^x>1$?

From $e^x > 1$ we get that $f'(x) > 0$ for $x>0$. Combine this with $f(0) = 0$, and we get $f(x)> 0$ for $x>0$.

how do we prove $f'(x)$ is positive/negative over an interval $[a,b]$?

That really depends on $f, a$ and $b$. In this case, we want to prove that $f'(x) = e^x-1$ is positive on the interval $(0, \infty)$, which comes down to showing that $e^x>1$. Where you go from here depends very much on your exact definition of $e^x$.

Answer 4

By definite integeration:

For $x>0$,

$$\int_0^xe^tdt>\int_0^xdt$$

$$e^x-1>x$$

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By Mean Value Theorem:

For $x>0$, there exists $\zeta\in(0,x)$ such that

$$\frac{e^x-e^0}{x}=\left.\frac{de^t}{dt}\right|_{t=\zeta}=e^\zeta>1$$

Answer 5

I believe you are over-complicating it. $e^x$ is positive and log-convex, hence convex. In particular its graph lies above any tangent line, like, for instance, the tangent line at $(0,1)$, whose equation is $y=x+1$.

Answer 6

Knowing that $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n,$$

Then you can define $e_n^x=\left(1+\frac{x}{n}\right)^n.$

It is easy to show that $e_2^x>x+1$ for $x>1,$ and using induction demonstrate $e_n^x>x+1,$ for $n\geq 2,$ giving you

$$e^x=\lim_{n\to\infty}e_n^x>x+1,$$

as desired.