# Show that polynomial has one real root

• Sep 8th 2010, 04:59 AM
Glitch
Show that polynomial has one real root
The question:
Show that the polynomial p, where $p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$ has at least one real root.

I'm not sure what to do. Solving for 0 is difficult with a cubic, so I don't think this is the approach my text is looking for. This question is within some maxima-minima problems, so I'm thinking I need to use that train of thought. As far as I can tell, the derivative of p has no real roots, so I'm still stuck.

Any suggestions? Thanks.
• Sep 8th 2010, 05:04 AM
Prove It
The simplest way would be to substitute some values for $x$ and show that you get both positive and negative values for $f(x)$. This means that $f(x)$ must have changed sign somewhere, and thus must have crossed the $x$ axis somewhere.
• Sep 8th 2010, 05:16 AM
Glitch
Indeed. However, I suspect that we're expected to be more rigorous in our approach.
• Sep 8th 2010, 05:19 AM
mr fantastic
Quote:

Originally Posted by Glitch
Indeed. However, I suspect that we're expected to be more rigorous in our approach.

The suggestion given IS rigorous. Review the Intermediate Value Theorem.
• Sep 8th 2010, 05:24 AM
Glitch
I know the IVT. It just didn't cross my mind that it would support Prove It's solution. Thanks guys.
• Sep 8th 2010, 10:25 AM
CaptainBlack
Quote:

Originally Posted by Glitch
The question:
Show that the polynomial p, where $p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$ has at least one real root.

I'm not sure what to do. Solving for 0 is difficult with a cubic, so I don't think this is the approach my text is looking for. This question is within some maxima-minima problems, so I'm thinking I need to use that train of thought. As far as I can tell, the derivative of p has no real roots, so I'm still stuck.

Any suggestions? Thanks.

A polynomial with real coefficients has complex roots that occur in conjugate pairs. Therefore a polynomial of odd order has an odd number of roots but an even number of complex roots and so must have at least one real root.

CB
• Sep 8th 2010, 11:08 AM
Quote:

Originally Posted by Glitch
The question:
Show that the polynomial p, where $p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$ has at least one real root.

I'm not sure what to do. Solving for 0 is difficult with a cubic, so I don't think this is the approach my text is looking for.

This question is within some maxima-minima problems,

so I'm thinking I need to use that train of thought. As far as I can tell, the derivative of p has no real roots, so I'm still stuck.

Any suggestions? Thanks.

You can find local maxima and minima by equating the derivative to zero.

$\displaystyle\ f(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$

$\displaystyle\ f'(x)=\frac{x^2}{2}+x+1$

The derivative being zero has complex solutions as $\displaystyle\ f'(x)=0\Rightarrow\ x=\frac{-1\pm\sqrt{1-2}}{1}$

hence there are no local maxima or minima.

Also notice that the derivative is always positive, so the graph is constantly increasing.
To see this, we can examine for which x gives the least value for the derivative

$\displaystyle\ f'(x)=\frac{x^2}{2}+x+1$

$f''(x)=x+1$

This is zero, giving least value for the derivative, when $x=-1$

$\displaystyle\ x=-1\Rightarrow\ f'(x)=\frac{1}{2}$

The slope of the tangent never goes below 0.5 (the graph is always increasing).

From here, you can again consider Prove It's statements, as $f(0)=1$

so the graph crosses the x-axis only once to the left of that.
This helps show that the polynomial has only one real root,