Warm up problem 19: undefined function
x 4 + 3 x 3 − 8 x 2 − 5 x 8 + 6 x 4 + 3 x 3 − 1 4 x + 6
For how many real values of x is the above expression undefined?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Alright, here's the calculus solution. Can someone post the algebraic one... If it exists (I swear there must be one)
We are only looking at the denominator an when it equals 0. Thus since it is not factorable, we can take its first derivative, set it equal to 0, and find its local maxima and minima.
Thus we have f ′ ( x ) = 4 x 3 + 9 x 2 − 1 6 x
The first obvious root is x=0, thus we check f ( 0 )
f ( 0 ) = ( 0 ) 4 + 3 ( 0 ) 3 − 8 ( 0 ) 2 − 5 ⇒ − 5
Since f(0) is less than 0, the leading coefficient is positive, and the function is of an even degree, we can conclude that f(x) has a maximum of two foots.
Through some plug and chug, we find that f ′ ( x ) has two other roots, one which is less than zero and the other which is greater, thus f ( 0 ) is the middle and only local maxima, thus the other two roots of f ′ ( x ) must be relative minima. Therefore, since f ( x ) increases as |x| increases towards a large enough quantity, f(x) has two roots and thus our original function is undefined at 2 points.
How about complex roots?
Log in to reply
Yes, I changed the wording. So if you put four, you should get credit
If you insists of going for the calculus route:
You can stop at "Thus we have f ′ ( x ) = 4 x 3 + 9 x 2 − 1 6 x " and show that if f ( x ) has four real roots. then f ′ ( x ) = 0 has 2 positive roots and 1 negative root, which is clearly false. This leaves us with either f ( x ) has no real roots or 2 real roots. If f ( x ) has 0 real roots, then f ( x ) must be strictly positive, which also clearly false because f ( 0 ) < 0 . Hence, the only possible solution is 2 real roots.
So we just want to make the denominator zero. Let f ( x ) denote the polynomial in the denominator. By Descartes Rule of Sign, we have one change in sign of f ( x ) and one change in sign of f ( − x ) thus there is exactly two roots. Hence the answer is 2.