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The question is that find all numbers $x$ such that $x+3^x <4$.

This is a question from problem $4$ from Spivak Calculus book,

My attempt:

It easy to see that $1$ is a solution of this equation $f(x)=x+3^x -4=0$, and$ f'(x)=1+\ln(3)e^{x\ln(3)}>0$ $ \forall x \in \mathbb{R}$

so $f(x)$ is strictly increasing. So $ \forall (x,y) \in \mathbb{R}$ $x<y $$\Rightarrow $$ f(x)<f(y)$

and we have $x+3^x <4$$\Rightarrow $$f(x)<0$$\Rightarrow$$f(x)<f(1)$

Now if we can show that $f(x)$ is bijective we can say that $f(x)<f(1)$$\Rightarrow $$x<1$.

i have two question :

$(q1)$ does my attempt correct?

$(q2)$

I am a student in the second year of a university majoring in mathematics and i do self-learning of calculus through "Spivak calculus", after finishing the first chapter of the book ,I am confused how I should solve the exercises 'Should I rely only on the previous paragraph or what !!'

the above problem exist at the first chapter(basic properties of numbers) of the book ,so we must to solve it without using some thing like bijective and continuous function...,??!!

Robert Z
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    You don't need to show if it is or not bijective. You have shown that $f$ is strictly increasing and that $f(1)=0$, then $1$ is the only zero of $f$. Now, as $f(0) <0$ and $f(2) >0$, by the continuity of $f$, we know that $f(x)<0$ iff $x<1$. That's your solution. – Eparoh Mar 12 '22 at 10:17
  • yeah ,i think i must just add the function is continuous in my solution ,because function increasing and continuous , is a bejictive function so $f(x)<f(1)$$\Rightarrow $$x<1$? –  Mar 12 '22 at 10:22
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    Thinking of it, you don't really need continuity nor bijectivity. Because if it is strictly increasing, then if $x<1$ then $f(x) < f(1) = 0$ and if $x>1$ then $f(x)>f(1)=0$, so you get that $f(x)<0$ iff $x <1$. And, you can also prove it is increasing without using derivatives, so you only need the machinery of the first chapter. – Eparoh Mar 12 '22 at 10:26
  • @ Eparoh thank you so much –  Mar 12 '22 at 10:30
  • @Eparoh We need the continuity : $-\frac{1}{x}$ is strictly increasing as well, but we do not have $x_1<x_2\implies -\frac{1}{x_1}<-\frac{1}{x_2}$ Counterexample is , for example , $x_1=-1$ , $x_2=1$. What we do not need is the bijectivity. – Peter Mar 12 '22 at 10:47
  • @Peter But then, by definition, the function $-\frac{1}{x}$ is not strictly increasing in $\mathbb{R}$. It is strictly increasing in $(-\infty, 0)$ and in $(0,\infty)$, but not in $\mathbb{R}$ nor in its domain. – Eparoh Mar 12 '22 at 10:53
  • @Peter Nevertheless, $-\frac{1}{x}$ is continuous in its domain. – Eparoh Mar 12 '22 at 10:59
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    https://math.stackexchange.com/a/4202342/754927 – Ben Mar 13 '22 at 15:12
  • @ben thank you ,after i saw you account i think you are already study throught spivak calculus book, did you have any advices for self learning through it? –  Mar 13 '22 at 16:04
  • @A.CH I'm only about halfway through and I've had to stop for a few months so I'm quite rusty. Here are past posts related to Spivak and self study in general. Hopefully they are helpful. https://math.stackexchange.com/a/3866993/754927

    https://math.stackexchange.com/a/4142526/754927

    https://math.stackexchange.com/a/4073563/754927

     – Ben Mar 13 '22 at 20:21
  • @ben,again thank you so much. –  Mar 14 '22 at 04:43
  • You're very welcome. Spivak is a grand adventure! – Ben Mar 14 '22 at 13:55

2 Answers2

1

Once you show that the function is strictly increasing, you're almost done. You just need to find $x_0$ such that $f(x_0)=4$ and you'll have that

  • for $x<x_0$, $f(x)<4$;
  • for $x>x_0$, $f(x)>4$.

The function $f\colon\mathbb{R}\to\mathbb{R}$ is indeed bijective, but this is not required for solving the problem.

For instance, $g(x)=e^x$ is increasing as well and you have $g(x)<1$ if and only if $x<0$, because $g(0)=1$. On the other hand, $g$ is not bijective, so you see this is unneeded for solving the similar problem.

egreg
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Option:

1)$x\le 0$ satisfies the inequality.

2)$x>0$;

$f(x):=x+3^x$ is stricly increasing (f is the sum of the two increasing functions $x$ and $3^x$).

By inspection $f(1)=4;$ satisfies the inequality.

Hence $x<1$ is the solution set.

Peter Szilas
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