Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$

Question

Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$.

This question is supposed to be solved without using derivatives.

Answer 1

The function is symmetric around the center point $x=\frac{7}{2}$. Letting $x = y + \frac{7}{2}$, we have:

$$ f(x) = \left| \left(y + \frac{7}{2}\right) \left(y + \frac{5}{2}\right) \left(y + \frac{3}{2}\right) \left(y + \frac{1}{2}\right) \left(y - \frac{1}{2}\right) \left(y - \frac{3}{2}\right) \left(y - \frac{5}{2}\right) \left(y - \frac{7}{2}\right) \right| $$

$$ f(x) = \left| \left(y^2-\frac{7^2}{4}\right) \left(y^2-\frac{5^2}{4}\right) \left(y^2-\frac{3^2}{4}\right) \left(y^2-\frac{1^2}{4}\right) \right| $$

Since $3 \leq x \leq 4$, $-\frac{1}{2} \leq y \leq \frac{1}{2}$ and $y^2 \leq \frac{1}{4}$, so all four factors are non-positive.

$$ f(x) = \left(\frac{7^2}{4} - y^2\right) \left(\frac{5^2}{4} - y^2\right) \left(\frac{3^2}{4} - y^2\right) \left(\frac{1^2}{4} - y^2\right) $$

$f(x)$ clearly decreases when $y^2$ increases, so the maximum of $f$ is achieved when $y^2$ is as small as possible. The smallest possible is $y^2=0$, at $x=\frac{7}{2}$.

$$ \max_{3 \leq x \leq 4} f(x) = \frac{7^2 \cdot 5^2 \cdot 3^2}{2^8} = \frac{11025}{256} \approx 43.066 $$

Answer 2

The maximum of the product occurs at the maximum of the logarithm of the product. Write this out to see the maximum must occur at $x = 3.5$.

Specifically, maximize $\log x + \log (x-1) + \cdots + \log (x -7)$

Consider the sum of just the first and last terms. How would you maximize that sum? (Convince yourself that it would be for $x = 3.5$.) Now consider the sum of the second and the second-to-last term. Again, $x=3.5$. And so on.

Put these answers together to see the result.

Answer 3

Graphical Comment. The function crosses the x-axis at $0, 1, 2, \dots, 7.$ So it seems that the maximum for $x$ in $(3,4)$ must be about halfway between $3$ and $4.$

curve(abs(x*(x-1)*(x-2)*(x-3)*(x-4)*(x-5)*(x-6)*(x-7)), 3, 4, ylab="y", main="")

A numerical search shows that the maximum $y= 43.0664$ is very nearly at $x = 3.5.$

x= seq(3,4,by=.001)
y = abs(x*(x-1)*(x-2)*(x-3)*(x-4)*(x-5)*(x-6)*(x-7))
x[y==max(y)]
[1] 3.5
max(y)
[1] 43.06641

Computations in R.

Answer 4

If the use of AM-GM inequality is allowed, then $$ f(x)=x(7-x) \cdot (x-1)(6-x) \cdot (x-2) (5-x) \cdot (x-3)(4-x) \\ \le \left(\frac 72\right)^2 \cdot \left(\frac 52\right)^2 \cdot \left(\frac 32\right)^2 \cdot \left(\frac 12\right)^2 = \frac{11025}{256}. $$ And all four equality hold simultaneously at $x=3.5$.