# How to find number of roots of a 5th degree polynomial using monotonicity?

Q) Let $$a \in R$$ and let $$f: R \rightarrow R$$ be given by $$f(x) = x^5 - 5x + a$$. Then -

(A) $$f(x)$$ has three real roots if $$a>4$$

(B) $$f(x)$$ has only one real root if $$a<-4$$

(C) $$f(x)$$ has three real roots if $$a<-4$$

(D) $$f(x)$$ has three real roots if $$-4

In this question, I first differentiated $$f(x)$$ and named it another function $$g(x)$$.

$$f'(x) = g(x) = 5x^4 - 5$$

Also, $$g(0) = -5$$, which means it has two roots.

$$\implies g(x)$$ is strictly decreasing in $$(-\infty, -5]$$ and strictly increasing in $$(5, \infty)$$.

I am stuck now. I don't understand how to proceed from here. I request to whoever answering this question, to tell me how to solve this using the method I am following. I am well aware that this question can be solved in other easier ways too. However, I want to solve this using ONLY this method. Thank you.

You know that $$f$$ is strictly increasing on $$(-\infty,-1]$$ and on $$[1,\infty)$$ and strictly decreasing on $$[-1,1]$$. So
1. If $$a>4$$, then, since $$f(1)=a-4>0$$, $$f$$ has only one real root.
2. If $$a<-4$$, then, since $$f(-1)=a+4<0$$, $$f$$ has only one real root.
3. If $$-4, then, since $$f(1)<0$$ and $$f(-1)>0$$, $$f$$ has three real roots.