Proving the Existence of a Real Zero Point for a Function with a Given Interval

[mariama1]

Homework Statement

Show that this function $$f_{n}(x)= x^{5}+nx-1$$
has exactly one real zero point and it is in the interval
$$\left(\frac{1}{n+1},\frac{1}{n}\right)$$

Homework Equations

By calling the zero point $$a_{n}$$
decide if the series $$\sum \left(-1\right)^{n-1} a_{n}$$
converges absolutly or conditionally ??
For which x converge the power series $$\sum a_{n}x^{n}?$$

The Attempt at a Solution

I tried to substitute the two end points of the interval for the x in the function by (intermediate value theorem)
to show that we have exactly one zero point , is it useful to use this way ??

 

[vela]

You need to describe your argument a bit more. You may be able to use the IVM to show there's at least one zero in the interval, but that doesn't mean there's only one zero in the interval or that there aren't zeros outside the interval. You'd still have to show that f_n(x) has exactly one zero.

 

[mariama1]

You need to describe your argument a bit more. You may be able to use the IVM to show there's at least one zero in the interval, but that doesn't mean there's only one zero in the interval or that there aren't zeros outside the interval. You'd still have to show that f_n(x) has exactly one zero.

and how can i show that ? can you help me ?

 

[lurflurf]

If f monotone?

 

[mariama1]

If f monotone?

Sorry,but i do not know :(

 

[vela]

Since f_n(x) is a polynomial, it's continuous everywhere. If it has more than one zero, it has to cross the x-axis, turn around, and cross the axis again. How can you show that it doesn't turn around?

 

[mariama1]

Since f_n(x) is a polynomial, it's continuous everywhere. If it has more than one zero, it has to cross the x-axis, turn around, and cross the axis again. How can you show that it doesn't turn around?

Yessss, I understod now thanks a lot for your good advices .
I solved it