Find the number for real roots of the function $f(x)= x^5-5x+a$ .

Question

Let $a\in R$ and $f:R\rightarrow R$ be given by $f(x)=x^5-5x+a$ then which of the following is true?

a) f(x) has 3 real roots if $a>4$

b) f(x) has 1 real root if $a>4$

c) f(x) has 3 real roots if $a<-4$

d) f(x) has 3 real roots if $-4<a<4$

I tried to proceed by using Descrates' Rule of Signs.

For $a>4$ a is definitely positive. So f(x) has 2 sign changes hence two positive real roots and f(-x) has 1 sign change hence one negative real root.So for $a>4$ i get f(x) has 3 real roots but the answer appears to be wrong. Similarly i tried for all four option and couldn't get the correct answer.

I want to know where i am doing mistake, can this rule be used here and how ? , how to solve this using application of derivatives.Please help. Thank you.

Answer 1

If $a > 0$, $f(x)$ has two sign changes, hence Descartes says there are either $2$ or $0$ positive real roots (counted by multiplicity).
Hint: $f'(x) = 5 x^4 - 5 = 0$ for $x = 1$, with $f''(1) = 20 > 0$, so $f$ has a local minimum at $x=1$, and the sign of $f(1) = a- 4$ will determine how many positive roots there are.