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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.

Can someone please help me prove this fact?

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    $\begingroup$ The second statement is false. Consider $f(x) = x^2+1$. $f'(x) =x$ which has a single root, but $f$ has no root. $\endgroup$ – msm Oct 22 '18 at 3:39
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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 $x_0<x_1$ .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 $f'(x)>0$

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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.

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

$f(0)=-6$ 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 $f^{\prime}(x)=12x^2+4>0\ \forall x\ne0$

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    $\begingroup$ 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. $\endgroup$ – P Vanchinathan Oct 22 '18 at 4:00
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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:

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

A contradiction.

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

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