Working on problem 11 in http://www.sosmath.com/cyberexam/precalc/TA1001/TA1001.html
On what interval(s) is the function f(x) = 7x - x3 increasing?
I determined that the x-intercepts are 0, +√7, and -√7.
So far so good.
I create a sign chart
[-∞, -√7] Positive
[-√7, 0] Negative
(0, √7] Positive
[√7, ∞] Negative
I determined that the graph looks something like this
At this point I realize, what I need to do is calculate the local minimum between -√7 and 0, as well as the local maximum between 0 and √7.
Since this is a pre-calculus question, I cannot resort to taking a derivative.
If this were a quadratic equation, I could calculate the vertex, but it's not.
So I'm stuck. How does one calculate the local maximum and minimum in this case?
Best Answer
The function is symmetric, so if a is the value where f(x) reaches its maximum on $(0, \sqrt 7)$ then the minimum occurs at -a.
One special quality the max has is that except for x = a if $ 0 \le x_1 \le \sqrt 7$ there is an $x_2$ in that interval such that $f(x_1) = f(x_2)$. That would give
$7x_1 - x_1^3 = 7x_2 - x_2^3$
$7(x_1-x_2) = x_1^3 - x_2^3$ or finally
$7 = x_1^2 + x_1x_2 + x_2^2$.
However at the max point a, $x_1$ has to be the same as $x_2$ and that last equation gives:
$3a^2 = 7$ and $a = \pm \sqrt(7/3)$.
And guess what -- that checks out with the calculus solution.