Find the number of roots lie in interval 
Let $a \in R $ and let $f : R \rightarrow R $ be given by $f(x)=x^5 -5x + a $
Then 
  
  
*
  
*$f(x)$ has three real roots if $a \gt 4$
  
*$f(x)$ has only one real roots if $a \gt 4$
  
*$f(x)$ has three real roots if $a \lt 4$
  
*$f(x)$ has three real roots if $ -4 \lt  a \lt
 4$

My work
If $$f(x)=0$$
$$ a=5x-x^5 =g(x)$$
$$g(x)=5x-x^5=0$$
$$x=0,5^{\frac{1}{4}},-5^{\frac{1}{4}}$$
I actually don't know how to do this type of question . Please tell me how to solve this type of question .
 A: Look at the stationary points, their $x$-coordinates satisfy $x^4 =1 \iff x = \pm 1$ (since we are working over the reals. Now check $f(1) = a-4$ (minimum) and $f(-1) = a+4$ (maximum). Plotting the function for $a=0$ to develop our intuition of the problem is a good idea at this stage: 

As can be see, the function will cross the $x$-axis at three distinct points as long as the maximum value of the function is greater than $0$, i.e: $a + 4 > 0 \iff a > -4$ and the minimum value of the functions is smaller than $0$, that is $a - 4 < 0 \iff a < 4$. So the required conditions on $a$ for there to be three roots are $$a \in (-4, 4)$$
I shall leave it to you to formalise the argument to your hearts content if you wish as well as investigate other cases for which intervals $a$ lies in, you can find restrictions on $a$ for there to be one root, two roots, no roots, etc...
A: Hint:
Using the derivatives:
$$
y'=5x^4-5 \qquad y''=20x^3
$$
you can see that the functions has a local maximum for $x=-2$ with value $y(-1)=a+4$ and a minimum for $x=1$ with value $y(1)=a-4$
Now note that, since the function is a $5-$degree polynomial it has always at least a real root, and the limit of the function for $x\to -\infty$ is $-\infty$ and the limit of the function for $x\to +\infty$ is $+\infty$ 
So, if $y(-1)$ and $y(1)$ have the same sign, there is no roots between $-1$ and $1$ and the function is alway negative before the unique root and positive after it.
If $y(-1)$ and $y(1)$ have opposite signs, than there is a root between them and other two roots, one before $-1$ and the other after $1$.
Finally you have the possibility that $y(-1)=0$ or $y(1)=0$, and , in this case, the function has a simple root and a double root.
A: The roots are such that
$$x^5-5x=-a$$
which means that they are the intersection of the curve $y=x^5-5x$ with the horizontal $y=-a$.
By a study of the curve (extrema), you can establish the following table of variations
$$\begin{matrix}x&-\infty&&-1&&+1&&+\infty\\x^5-5x&-\infty&\nearrow&+4&\searrow&-4&\nearrow&+\infty\end{matrix}$$
This tells you that for $-a<-4$ or $-a>+4$ (below or above the two extrema), there is a single root, and there are three in between.

A: Considering the derivative $f'(x)=5(x^2-1)$, you see $f$ a a local maximum M=f(-1) first, then a local minimum $m=f(1)$.  Thus,


*

*$f(x)$ has one real root if $m$ and $M$ have the same sign, i.e. if and only if 
$$mM=f(1)f(-1)=(a-4)(a+4)>0\iff\begin{cases}a>4 \enspace\text{or}\enspace a<-4\end{cases},$$

*$f(x)$ has three roots if $m$ and $M$ have different signs, i.e. if and only if $-4<a<4$,

*$f(x)$ has one multiple root and one simple root if $a=\pm4$.  One checks the multiple root ($1$ or $-1$) is a double root.

