To find number of real roots Consider the equation $x^5-5x=c$ where c is a real number.
Determine all c such that this equation has exactly 3 real roots.
I know that between consecutive real roots of $f$ there is a real root of $f'$.
Now $f'$ in this case is $5x^4-5$ which always has two real roots. So the claim should be true for all c.
But I KNOW IT IS NOT TRUE. Where am I messing up? 
 A: Yes, between any two roots of $f$, there is a root of $f'$. However, just because $f'$ has a root, that doesn't mean that $f$ has a root on either side. Consider $f(x)=x^2+1$.
As for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $c$ we have three roots, for some values of $c$ we have a single root, and for exactly two values of $c$, there are two roots. The three-root region is exactly the interval between the two two-root values of $c$.
And finding the values of $c$ that gives two roots is easier than one might think. They happen exactly when one root of $f$ coincides with a root of $f'$. So find the roots of $f'$, and find the values of $c$ that make each of them a root of $f$, and you have found the interval of $c$-values that gives three roots.
A: As you know, the given equation has extrema at $x=\pm1$. These correspond to values of the polynomial
$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left).
Hence the polynomial will grow from $-\infty$, reach the maximum, then the minimum and continue growing to $\infty$. There are three roots when $0$ is in the range $(-4-c,6-c)$.
A: Where am I messing up?  Just look at a graph where it fails.
$$
x^5-5 x -5
$$

A: 
I know that between consecutive real roots of f there is a real root of f′. Now f′ in this case is 5x4−5 which always has two real roots. So the claim should be true for all c.

$A \implies B$ does not mean $B \implies A$. 
The two real roots of $5x^4 - 5$ are the two  roots at $x = \pm 1$.
If $x^5 - 5x=c$ has three roots then they will be at $x < -1; -1 < x < 1; $ and at $x > 1$ by your condition. 
But there won't be three real roots if there is no root for any $x< -1$, or no root between $-1$ and $1$, or no root for any $x < -1$. 
$x=\pm 1$ are extreme points and if one, the max, is $>0$ and the other $<0$ then there will be three real roots.  But if both are "on the same side of $0$" there is no root between them and no root to "the other side".
$x^5-5x -c|_{-1} = 4-c$ and $x^5 - 5x -c|1 = -4-c$ so $x =-1$ is a max and $x = 1$ is a min.
If $f(-1) = 4-c \le 0$ is a max there will be no root for $x < -1$ or for $-1 < x \le 1$.  If $f(1) = -4-c\ge 0$ is a min there will be no root for $x > 1$ or for $-1 \le x < 1$.  So if either $c \ge 4$ or if $c\le 4$ then there are fewer than three real roots.  But if $-4 < c < 4$ then there will be three.
Alternatively:  we know what the shape of an odd  polynomial $x^5 -5x$ looks like.  It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $x \to \infty$, goes to negative infinity as $x \to -\infty$.  It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative).  If we shift it up or shift it down by $c$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root.  Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots.  So if $c$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits.  If $c$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root.
