Thread: Let P(x) = x^5 + ax +b

Originally Posted by Susie38

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.

I am going to use the root-location theorem.

If is countinous on and have opposite signs then there exists a number such as, .

You have, thus,

Find the critical points,
Thus, but since we have, thus there are no critical points (because it yields imaginary numbers). Because 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

Because the function is always increasing. Meaning it intersects the x-axis only one

Actually that is not necessarily true. It would have been correct to state that since,


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.

Originally Posted by Susie38

Let p(x) = x^5 + ax + b where a and b are real numbers.
...
b.) What is the maximum number of real roots p(x) may have? Justify your answer.

Hello,

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 .

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].