Problem: Prove that the function
f (x) = x4 + 6x2 + 24x − 13
has exactly two roots.
Solution: We will prove this assertion in two steps. First, we prove that there cannot be more than two roots. Second, we will show that we actually do have two.
1. To prove that we cannot have more than two roots we will use one of the corollaries to Rolle's theorem, see the section Tricks with MVT in Theory - MVT.
By contradiction, assume that we have three or more roots. By the Fact mentioned above, the first derivative would have to have two or more roots. The derivative is
f ′(x) = 4x3 + 12x + 24
and as a polynomial of degree three it might conceivably have two or more roots, without further work we cannot learn more. Instead of investigating this first derivative we apply the Fact once more.
Assuming that the function f has three or more roots, the first derivative would have to have two or more and the second derivative would consequently have to have one or more roots. However, the second derivative is
f ′′(x) = 12x2 + 12
and it obviously does not have any root, a contradiction.
This proves that we can have at most two roots.
2. Now we will show that we actually do have two roots. While there are formulas for finding roots of a fourth degree polynomial, we prefer not to use them, since they are ugly and we do not remember them anyway. Instead, we will try a more general approach that has the advantage that it has a chance to work also for polynomials of higher degree.
We will use the Intermediate value theorem (see Continuity in Functions - Theory - Real functions). We will start substituting nice numbers into f, waiting for the values to change signs. We start with 0 and start going in positive and negative direction by 1.
| x: | −3 | −2 | −1 | 0 | 1 |
| f (x): | 50 | −21 | −30 | −13 | 18 |
| sign: | + | − | − | − | + |
Since the function f is continuous and it changes its sign at two
places, we conclude using the Intermediate value theorem that the function
has a zero point in the interval
Remark: We can use the bisection method (see Sequences - Theory - Applications) to narrow down these two roots. We start with the first root.
f (−2.5) = 3.5625 > 0, thus the root is between
−2.5 and −2.
f (−2.4) = −2.8624 < 0, thus the root is between
−2.5 and −2.4.
f (−2.45) = 0.24500625 > 0, thus the root is
between −2.45 and −2.4.
Now we similarly look at the second root.
f (0.5) = 0.5625 > 0, thus the root is between
0 and 0.5.
f (0.4) = −2.4144 < 0, thus the root is
between 0.4 and 0.5.
f (0.45) = −1.91295625 < 0, thus the root is
between 0.45 and 0.5.
Analogously we could find these roots with any required precision.