# 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 $3 \leq x \leq 4$

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.

• Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Nov 30, 2020 at 18:54
• Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. Commented Nov 30, 2020 at 18:54
• I'd probably look at $t=x-3.5$. Commented Nov 30, 2020 at 19:01

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 $$x=\frac{7}{2}$$.

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

• I was going to say that ... using the symmetry is neat. It is in fact easy to show that there is just one turning point between each of the zeros [just noting as a fact] and the symmetry shows where this one must be. Commented Nov 30, 2020 at 19:18
• @MarkBennet Is it "easy to show" with a non-calculus approach? We can certainly notice the derivative of the function inside $|\cdot|$ is a 7th degree polynomial with a zero somewhere in each of 7 distinct intervals. Commented Nov 30, 2020 at 23:42
• Probably not without calculus thinking about it - fair comment. Commented Dec 1, 2020 at 6:09

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.

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 $$x=3.5$$.

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.