The simplest way would be to substitute some values for and show that you get both positive and negative values for . This means that must have changed sign somewhere, and thus must have crossed the axis somewhere.
The question:
Show that the polynomial p, where 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.
The derivative being zero has complex solutions as
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
This is zero, giving least value for the derivative, when
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
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.