The Art and Craft of Problem Solving

9.3 DIFFERENTIATION AND INTEGRATION 329

that at relative minima, the second derivative is non-negative. Thus p"(a) 2 o. But
p(a) 2 p"(a) by hypothesis, which contradicts p(a) < O. •

Here is another polynomial example which adds analysis of the derivative to stan­
dard polynomial techniques.

Example 9.3.2 (United Kingdom 1995) Let a, b, e be real numbers satisfying a < b <

e, a + b + e = 6 and ab + be + ea = 9. Prove that 0 < a < 1 < b < 3 < e < 4.

Solution: We are given information about a + b + e and ab + be + ea, which sug­
gests that we look at the monic cubic polynomial P(x) whose zeros are a, b, e. We
have

P(x) = x^3 - 6.x2 + 9x - k,

where k = abe, using the relationship between zeros and coefficients (see page 168).
We must investigate the zeros of P(x), and to this end we draw a ro ugh sketch of the
graph of this function. The graph of P(x) must look something like the following
picture.

----f----------:-..-----'---+---- x

We have not included the y-axis, because we are not yet sure of the signs of a, b, e.
But what we are sure of is that for sufficiently large negative values of x, P(x) will
be negative, since the leading term x^3 has this behavior, and it dominates the other
terms of P(x) if x is a large enough negative number. Likewise, for sufficiently large
positive x, P(x) will be positive. Since the zeros of P(x) are a < b < e, P(x) will
have to be positive for x-values between a and b, with a relative maximum at some
point U E (a,b) , and P(x) will attain negative values when b < x < e, with a relative
minimum at some point v E (b,e).
We can find u and v by computing the derivative
P'(x) = 3.x2 - 12x+9 = 3(x - 1)(x - 3).

Thus u = 1, v = 3 and we have /(1) > 0, /(3) < 0, so a < 1 < b < 3 < e.
It remains to show that a > 0 and e < 4. To do so, all we need to show is that
P(O) < 0 and P( 4 ) > O. We will be able to determine the signs of these quantities if we

can discover more about the unknown quantity k. But this is easy: P( 1) = 4 - k > 0 and

P( 3 ) = -k < 0, so O < k < 4. Therefore we haveP(O) = -k < 0 and P( 4 ) = 4 -k > 0,