Real Roots of x^5 + x + c Equation: [-1,1]

[lkh1986]

Question: How many real roots does the equation x^5 + x + c = 0 have on the interval of [-1,1]?

I try to differentiate the equation, then I obtain 5x^4 + 1.

When at its local minimum or local maximum, 5x^4 + 1 = 0.

So, there is no solution for the equation, since 5x^4 + 1 > or = 1

So, I conclude that the graph of x^5 + x + c is increasing when x increases. So, I think the answer to this question is there is at most 1 root for the equation x^5 + x + c = 0. (Since c can take any value, we can shift up and down the graph, so there is at least one real root)

Is my answer correct? Or is there any other way to solve thi problem? Thanks

 

[Galileo]

You are correct to say the function is nondecreasing everywhere, but you have to find the number of real roots on the interval [-1,1]. This will depend on the value of c.

 

[lkh1986]

Yes, so what I can know from here is that the graph increases all the way up. So, there is no local minimum or maximum. It is more or less like the graph of y = x^3, where it is a one-to-one function.

So, I assume that the graph of y = x^5 + x + c is also a one-to-one function, where therefore, the graph can only intercept x-axis at most one point.