Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I solved graphically and found that $x + 3^x < 4$ is true for $x < 1$ but I can't find a way to prove it algebraiclly, any hints will be greatly appreciated!

share|improve this question

3 Answers 3

up vote 4 down vote accepted

The left hand side is an increasing function of $x$. With $x=1$, its value is $4$.

share|improve this answer
    
I suppose we just take that $3^x$ is strictly increasing from the fact that $3^x = e^{x ln(3)}$ and the properties of $e$, right? –  Edgar Sánchez Dec 9 '12 at 23:58
    
Yes, or with smaller teaspoons, that $3>1$ so that $\ln 3>0$, and the exponential function (i.e., base $e$) is strictly increasing. –  Harald Hanche-Olsen Dec 10 '12 at 8:48

The solution is $x<1$

Since $x+3^x$ is strictly increasing. Therefore for all $x<1$ we have $x+3^x<1+3^1=4$.

If $1\leq x$ we have $1+3^1\leq x+3^x$ because $x+3^x$ is increasing.

share|improve this answer
    
Thank you, that was fast! –  Edgar Sánchez Dec 9 '12 at 23:58

Having found the solution $1$ to $x+3^x=4$ toucan use the fact that the derivative is positive to show it is unique.

share|improve this answer
    
It's a pre-calculus problem so we are not allowed to use derivatives yet, but thank you anyway! –  Edgar Sánchez Dec 9 '12 at 23:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.