Why is $a=e$ the smallest number such that $a^x\ge 1+x$ for all $x$? [duplicate]

Calculus book:

Find all numbers $a$ such that $\forall x, a^x \ge 1+x$

I immediately thought of the inequality $e^x\ge 1+x$ and guessed that the answer was any number $a$ in $[e,\infty)$. After playing around with a graphing app I can see this is definitely true, although I can't explain why.

Why is $e$ specifically the smallest number such that the inequality holds? To my untrained eye, a quick scan of the equation does not give rise to anything involving the constant $e$. Maybe because $1+x$ is only a tangent to the equation $a^x$ when $a=e$? Thanks.

marked as duplicate by Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 29 '17 at 12:36

• $e^x$ is not the smallest number $\ge 1+x$. – Lord Shark the Unknown Nov 29 '17 at 5:50
• Have you worded your question right? Are you looking for smallest $a$ such that $a^x \geq 1+x , \, \forall x$? – baharampuri Nov 29 '17 at 6:00
• @baharampuri Actually, I'm looking for all numbers $a$ that satisfy that constraint for all $x$. The answer is probably in the form of an interval. – Lug Gian Nov 29 '17 at 6:21
• @LordSharktheUnknown Judging from your five upvotes on that comment, I've made a mistake in my question. But the two answers I have (including yours) confirm $e$ is the smallest number. Have I made a mathematical typo? You seem to be understanding my question. – Lug Gian Nov 29 '17 at 6:32
• @LugGian: I've now fixed the title so it actually fits your question. – celtschk Nov 29 '17 at 6:44

Suppose that $0<a<e$. The derivative of $f(x)=a^x$ is $f'(x)=(\ln a )a^x$. In particular $f'(0)=\ln a<1$. Since $\ln a<(1+\ln a)/2<1$, there is a $\delta>0$ such that $$f(x)<f(0)+\frac{1+\ln a}2x<1+x$$ when $1<x<1+\delta$.
• OP speaking: You said $f'(1)=\ln a$, but I think you mean $f'(0)$? – Lug Gian Nov 29 '17 at 6:28
Note that $e^x$ has taylor series: $$e^x = 1+x+x^2/2!+\dots$$ We can rewrite $a^x = e^{x\ln a}$, then this has taylor series: $$a^x = e^{x\ln a} = 1+x\ln a+ x^2(\ln a)^2/2!+\dots$$ Now, if we want $a^x\geq 1+x$, this means: $$1+x\ln a+x^2(\ln a)^2/2!+\dots \geq 1+x\implies x(\ln a-1)+x^2(\ln a)^2/2!+\dots\geq 0$$ All of the coefficients of the taylor polynomial on the left are positive if and only if $\ln a-1\geq 0\implies \ln a \geq 1$, so if $a\geq e$.
If $a<e$, then for $x>0$ the inequality won't hold. I'm guessing it wouldn't be too bad to compute specific intervals where the inequality fails, but I'll omit this for now.