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I have encountered this inequality in Spivak's Calculus (first chapter exercises), which I'm not sure how to solve.

$$ 3^x + x < 4 $$

I might be wrong but my gut feeling says the inequality holds for all $x$ between $(-\infty, 1)$ but I cannot prove it.

As I read, there seems to be no standard scheme for solving this type of inequalities/equations. How do You then usually proceed when dealing with one like the above? Thanks.

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Because $x$ and $3^x$ are increasing, so is $x+3^x$.

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  • $\begingroup$ Is it rigorous enough? $\endgroup$ – Eval Mar 26 '18 at 9:12
  • $\begingroup$ You should know it from the properties of elementary functions ($a^x$ is strictly increasing for $a>1$). Moreover, it looks as an exercise before introducing derivatives. Am I right? And you should not expect a general method. Changing the constatnt from 4 to 5 makes the exercise unsolvable in a simple way. $\endgroup$ – Przemysław Scherwentke Mar 26 '18 at 9:14
  • $\begingroup$ Yes, it is before Derivatives. Could you please explain the last point? Why would changing from 4 to 5 make it unsolvable in a simple way? $\endgroup$ – Eval Mar 26 '18 at 9:28
  • $\begingroup$ In the case 4 it is easy to guess that $x=1$ is the solution of equality. If you have 5, you can only estimate the solution numerically. $\endgroup$ – Przemysław Scherwentke Mar 26 '18 at 9:33
  • $\begingroup$ I see what you meant, thanks. $\endgroup$ – Eval Mar 26 '18 at 9:34
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defining $$f(x)=3^x+x-4$$ then we get $$f'(x)=3^x\ln(3)+1>0$$ can you finish?

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  • $\begingroup$ First chapter deals with basic properties of numbers and the Derivative comes much later. One is therefore supposed to solve it using basic properties of numbers There must be thus a solution involving other methods. $\endgroup$ – Eval Mar 26 '18 at 9:08

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