How to show that a cubic function is continuous, differentiable and has image $(-\infty, \infty)$? Suppose we have $f(x) = x^3 - 3x + 2$. How do I show that $f(x) = 5$ has a solution in a formal way?
I know that $f$ is continuous, $\displaystyle \lim_{x \to \infty} f(x) = \infty$ and $\displaystyle \lim_{x \to -\infty} f(x) = - \infty$ which implies, by IVT that $f(x) = 5$ has a solution.
But the use of limits that way and saying it continuous directly seems like a notation I haven't seen being used by authors as such.
 A: As noted by @geetha290krm, you can observe that $f(3)>5$ and $f(0)<5$. But notice that you don't have to explicitly find those specific points. Since you know that $\lim\limits_{x\to+\infty}f(x)=+\infty$, which means that $\forall M\in\mathbb{R}\ \exists k\in\mathbb{R}\ \forall x\geq k:f(x) > M$, you have that $\exists k\in\mathbb{R}\ \forall x\geq k:f(x)>5$ by taking $M=5$. Similarly, since you know $\lim\limits_{x\to-\infty}f(x)=-\infty$, you can find $\exists k'\in\mathbb{R}\ \forall x\leq k':f(x)<5$. Hence, since $f(k')<5$ and $f(k)>5$, you know by continuity that $\exists x\in(k',k):f(x)=5$. This is the formal reason why you can indeed apply the IVT on the limiting case (it is actually rigorous).
A: A constant function and the identity function are continuous. The sum and product of two continuous functions is continuous. (These properties are easy to prove from the first principles.) So any polynomial function is continuous.
As soon as you find two values of the polynomial on either side of $5$, you are done, differentiability is not required.
