Prove that the equation $ e^{x}+x^{3}=10+x $ has a unique solution on the open interval $(-\infty,\infty)$. Prove that the equation
$$
e^{x}+x^{3}=10+x
$$
has a unique solution on the open interval $(-\infty,\infty)$.

Letting $f(x)=e^{x}+x^{3}-10-x$, $f(0)&lt;0$ and $f(10)&gt;0$, 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 ($a&lt;b$ 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?
You can observe that if $x &lt;-1$ then $X^3-X &lt; 0$, thus
$$f(x) &lt;e^{-1}-10 &lt;-9$$

Also if $x \in [-1, 1]$ we have $x^3 \leq 1$ and $-x \leq 1$ thus 
$$f(x) \leq e+2-10 &lt; -5$$ 

Therefore there is no solution $x \leq 1$.

Now apply your technique on $(1, \infty)$. It is easy to show tha on this interval $f'(x) &gt;0$.
Prove that $f(x)$ is injective over positive real numbers. 

$$f'(x) &gt; 0, \forall x\geq0 \text{ appart for 0  and } f\in C^{\infty} (\mathbb{R}, \mathbb{R}) \implies f \text{ injective } $$ 

Then $\forall(a, b) \in \mathbb{R^+}^2, f(a) = f(b) \iff a=b$

And $\forall x&lt;0, f(x) &lt; 0$ as $x^3-x\leq1$ and $e^x&lt;1$