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!
Algebraic solution of x + 3^x < 4
2
$\begingroup$
inequality
3 Answers
4
The left hand side is an increasing function of $x$. With $x=1$, its value is $4$.
-
0Yes, or with smaller teaspoons, that 3>1 so that \ln 3>0, and the exponential function (i.e., base $e$) is strictly increasing. – 2012-12-10
2
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.
1
Having found the solution $1$ to $x+3^x=4$ toucan use the fact that the derivative is positive to show it is unique.
-
0It's a pre-calculus problem so we are not allowed to use derivatives yet, but thank you anyway! – 2012-12-09