Find roots of $3^x+x^3=17$ Find $x$ in the following equation:
$$3^x+x^3=17$$
 A: I will only give you a hint since this is a homework assignment. Try these numbers $-3, -2, -1, 0, 1, 2,3$. Are any of these a solution?
If you find one, try to show that this is the only solution. This you can do by considering the derivative of $f(x) = 3^x + x^3$:
$$
f'(x) = ...
$$
If you can show that this derivative is strictly positive or strictly negative, then you have shown that the function is strictly increasing (or strictly decreasing). Then it can only have one solution (you might have to think about why this would be true).
A: Or, compute the derivative and note that it is always $ > 0$. Then by Rolle's there is at most one solution. (It is easy to guess our solution...)
A: Note that if $y\gt 0$we have $3^y\gt 1$ so that if $x\ge 0$ $$3^{x+y}+(x+y)^3=3^x\cdot 3^y +x^3+(3x^2y+3xy^2+y^3)\gt 3^x+x^3$$
And it is very easy to show that for $x\le 0$ we have $3^x+x^3\le 1+0\lt 17$
So the function is increasing for positive $x$ (without calculus) and there is at most a single solution.
A: Please draw functions 3^x and 17-x^3.
By looking at the picture, you can see there is only one solution. Others have already proved that using basic computations.
Then you can bound your answer and check the value that makes this equation true is x=2.

