Show that polynomial has one real root

**Glitch** · September 8th 2010, 04:59 AM

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.

**Prove It** · September 8th 2010, 05:04 AM

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.

**Glitch** · September 8th 2010, 05:16 AM

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

**mr fantastic** · September 8th 2010, 05:19 AM

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.

**Glitch** · September 8th 2010, 05:24 AM

I know the IVT. It just didn't cross my mind that it would support Prove It's solution. Thanks guys.

**CaptainBlack** · September 8th 2010, 10:25 AM

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

**Archie Meade** · September 8th 2010, 11:08 AM

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,
in keeping with your title.

In the body of your question, you mention at least one real root.

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