Consider quartic function $y = f(x) = x^4 + ax^3 + bx^2 + cx + d \ (b, c, d ,e \in \mathbb R)$ which is illustrated by graph $(C)$. Knowing that $(C)$ is tangent to the $x$-axis at point with $x$-intercept $x_1$ and further intersects the $x$-axis at points with intercepts $x_2$ and $x_3$. Determine the number of points of intersection between the graph of the function $y = [f'(x)]^2 - f(x)f''(x)$ and the $x$-axis.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
I hope it's obvious but indeed, $$f(x) = (x - x_1)^2(x - x_2)(x - x_3) \iff \left\{ \begin{aligned} f'(x) &= (x - x_1)[4x^2 - (2x_1 + 3x_2 + 3x_3)x\\ &+ (x_1x_2 + x_1x_3 + 2x_2x_3)]\\ f''(x) &= 12x^2 - 6(2x_1 + x_2 + x_3)x\\ &+ 2(x_1^2 + 2x_1x_2 + 2x_1x_3 + x_2x_3) \end{aligned} \right.$$
It's evident that one of the roots of the equation $g(x) = 0$ is $x = x_1$, but are there any other ones? Of course, in principle, $\deg{g(x)} = 6$, which means there are a maximum of $6$ roots in the equation $g(x) = 0$.
Hmmm~ how about this? $$\begin{aligned} g(x) = [f'(x)]^2 - f(x)f''(x) &= -[f(x)]^2 \times \dfrac{f(x)f''(x) - [f'(x)]^2}{[f(x)]^2}\\ &= -[f(x)]^2 \times \dfrac{\mathrm d}{\mathrm dx}\dfrac{f'(x)}{f(x)} = -[f(x)]^2 \times \dfrac{\mathrm d^2}{\mathrm d^2x}\ln|f(x)| \end{aligned}$$
Actually, it's more like $\dfrac{f(x)f''(x) - [f'(x)]^2}{[f(x)]^2} = \dfrac{\mathrm d^2}{\mathrm d^2x}[\ln|f(x)| + C_1x + C_2]$, but we don't talk about that.
(Have you ever seen a busier graph?)
I'm currently in a rush, – =͟͟͞͞ =͟͟͞͞(¦3[=͟͟͞͞__]=͟͟͞͞=͟͟͞͞ =͟͟͞͞), so this is all you're getting, sorryyy~ As always, thanks for reading, (and even more so if you could help), have a great tomorrow, everyone~
By the way, the choices were $0, 1, 2$ and $3$.
Note that we can write
$$ g(x) = f(x)^2 (-\log | f(x) |)''. $$
Also, by noting that $f(x) = (x - x_1)^2(x - x_2)(x - x_3)$, we get
$$ (-\log |f(x)|)'' = \frac{2}{(x - x_1)^2} + \frac{1}{(x - x_2)^2} + \frac{1}{(x - x_2)^2}. $$
From this, we know that $(-\log|f(x)|)''$ never becomes zero (whenever it is defined) and hence $g(x) = 0$ holds only if $f(x) = 0$, i.e., $x \in \{x_1, x_2, x_3\}$. Moreover, plugging the above result to $g(x)$, we get
\begin{align*} g(x) &= 2(x-x_1)^2(x - x_2)^2(x - x_3)^2 + (x-x_1)^4(x - x_3)^2 + (x-x_1)^4(x - x_2)^2 \tag{*} \end{align*}
From this, it is clear that $g(x_1) = 0$ but $g(x_2) \neq 0$ and $g(x_3) \neq 0$. Therefore the answer is $\boxed{1}$.