Prove that the equation
$$
e^{x}+x^{3}=10+x
$$
has a unique solution on the open interval $(-\infty,\infty)$.
Letting $f(x)=e^{x}+x^{3}-10-x$, $f(0)<0$ and $f(10)>0$, so by the IVT there must be a solution in the interval $(0,10)$.
Now I must show there cannot be more than one solution. Generally for these proofs you assume that there is at least 2 solutions and show a contradiction. Letting the solutions be a and b ($a<b$ WLOG) I want to use Rolle's theorem to show that $f'(x)=e^{x}+3x^{2}-1$ cannot be zero (giving the contradiction). The problem is $f'(0)=0$. So how do I proceed?
Best Answer
You can observe that if $x <-1$ then $X^3-X < 0$, thus $$f(x) <e^{-1}-10 <-9$$
Also if $x \in [-1, 1]$ we have $x^3 \leq 1$ and $-x \leq 1$ thus $$f(x) \leq e+2-10 < -5$$
Therefore there is no solution $x \leq 1$.
Now apply your technique on $(1, \infty)$. It is easy to show tha on this interval $f'(x) >0$.