# How to prove $f(x) = 4x^{3} + 4x - 6$ has exactly one real root?

How can I show that $$f(x) = 4x^{3} + 4x - 6$$ has exactly one real root?

I think the best way is to show $$f'(x) = 12x^2 + 4 > 0$$ for all $$x \in \mathbb{R}$$. Thus, $$f'(x)$$ has zero real roots. Thus, $$f(x)$$ has at most one real root.

I thought about trying to show that if $$f$$ is a polynomial and $$f'$$ has $$n$$ real roots, then $$f$$ has $$n + 1$$ roots by using Rolle's Theorem or Mean Value Theorem, but I don't think this fact, in general, is true. I would need to prove this statement.

• The second statement is false. Consider $f(x) = x^2+1$. $f'(x) =x$ which has a single root, but $f$ has no root. – msm Oct 22 '18 at 3:39

Your intuition for the first one is correct!

$$f(0)<0$$ and $$f(1)>0$$, so by IVT $$f$$ has a root say $$x_0$$

Suppose $$f$$ has another root $$x_1 \neq x_0$$ with $$x_0 .Then $$f(x_0)=f(x_1)=0$$ and by Rolles theorem $$\exists$$ $$c \in (x_0,x_1)$$ such that $$f'(c)=0$$, contradicting to the fact $$f'(x)>0$$

Your $$f'(x)$$ is strictly positive which means your function is strictly increasing. A strictly increasing function does not have more than one real root. Because otherwise it is not going to be one-to-one. Simply put, with more than one real root you have to have a turning point somewhere between those roots which makes your function both increasing and decreasing between the roots.

$$f(x)=4x^3+4x-6$$

$$f(0)=-6$$ and $$f(-1)=-14$$

By IVT, there exists at least one real root, $$x\in(-1,0)$$ such that $$f(x)=0$$

Now try to prove by using contradiction.

If not, there exists al least $$2$$ real roots $$x_1,x_2$$, such that $$f(x_1)=f(x_2)=0$$

Since $$f(x)$$ is differentiable, by using Rolle's Theorem, there exists a number $$k\in(x_1,x_2)$$ such that $$f^{\prime}(k)=0$$. But $$f^{\prime}(x)=12x^2+4>0\ \forall x\ne0$$

• KIndly edit your answer: You have calculated and shown $f(x)$ takes negative values at both $x=0$ and $x=-1$. In this case IVT cannot allow us to conclude there is a root in between $-1$ and $0$. Possibly you mean $x=1$ where the function value is positive. – P Vanchinathan Oct 22 '18 at 4:00

Pedestrian:

$$f'(x)>0$$ implies $$f$$ is strictly increasing, i.e.for

$$x_1 < x_2$$ we have $$f(x_1) < f(x_2)$$.

Assume a strictly increasing function has more than one zero.

Let $$x_1 < x_2$$ be $$2$$ zeroes of the function:

$$f(x_1)=f(x_2)= 0$$.

P.S. Can there be a double zero if $$f'(x) >0$$?