Find all numbers $x$ such that $x+3^x <4$ ."spivak calculus"

Question

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…,??!!

Answer 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.