Let p(x) = x^5 + ax + b where a and b are real numbers.

a.) prove that p(x) has exactly one real root if a > 0.

b.) What is the maximum number of real roots p(x) may have? Justify your answer.

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- May 29th 2006, 01:54 PM #1

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- May 29th 2006, 04:16 PM #2

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Originally Posted by**Susie38**

If $\displaystyle f(x)$ is countinous on $\displaystyle [a,b]$ and $\displaystyle f(a),f(b)$ have opposite signs then there exists a number $\displaystyle a<x<b$ such as, $\displaystyle f(x)=0$.

You have, $\displaystyle f(x)=x^5+ax+b$ thus,

$\displaystyle f'(x)=5x^4+a$

Find the critical points,

$\displaystyle 5x^4+a=0$ Thus, $\displaystyle 5x^4=-a$ but since $\displaystyle a>0$ we have, $\displaystyle a<0$ thus there are no critical points (because it yields imaginary numbers). Because $\displaystyle f'(x)=5x^4+a>0$ the function is always increasing. Meaning that it intersects the x-axis only once. Thus, there is only one real root.

Originally Posted by**ThePerfectHacker**

$\displaystyle \lim_{x\to\infty}f(x)=\infty$

$\displaystyle \lim_{x\to-\infty}f(x)=-\infty$

We can make the function as large and as small as we like. Meaning that we can choose values such as it is negative and positive. Then apply the root location theorem.

- May 30th 2006, 12:06 PM #3Originally Posted by
**Susie38**

I presume that you have to answer this question if a, b are real numbers (without any restrictions).

By the first derivative (compare ThePerfectHacker's answer) you can only calculate 2 turning points**if a < 0**.

$\displaystyle x=-\sqrt[4]{\frac{-a}{5}}\ \vee \ x=\sqrt[4]{\frac{-a}{5}}$

Place the maximum over the x-axis and the minimum below the x-axis and you'll get a maximum number of 3 intercepting points with the x-axis, that means you get a maximum fo 3 roots.

I've attached a diagram with a family of functions:

The red graphs show what happens if the value of b has been changed.

The black graphs show what happens when the value of a has been changed. The domain of a and b is [-2,2].