Question

The number of points, at which the function \(f(x) = \max\{6x, 2 + 3x^2\} + |x - 1| \cos|x^2 - \frac{1}{4}|\), \(x \in (-\pi, \pi)\), is not differentiable, is ____.

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
The function \(f(x)\) is the sum of two parts: \(f_1(x) = \max \{6x, 2+3x^2\}\) and \(f_2(x) = |x-1| \cos \left| x^2 - \frac{1}{4} \right|\).
A function of the form \(\max\{g(x), h(x)\}\) is generally non-differentiable where the two constituent functions intersect, provided their slopes are different at the intersection points.
A function involving absolute values is typically non-differentiable where the argument of the absolute value becomes zero.
Step 2: Key Formula or Approach:
For \(f_1(x)\), solve \(6x = 2 + 3x^2\) to find the points where the function rule changes.
For \(f_2(x)\), examine the critical points where the terms inside absolute values become zero: \(x - 1 = 0\) and \(x^2 - \frac{1}{4} = 0\).
Check if the product with the cosine term smoothens out any non-differentiability.
Step 3: Detailed Explanation:
Let's analyze \(f_1(x)\):
Set the inner functions equal: \(3x^2 - 6x + 2 = 0\).
Using the quadratic formula: \(x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{1}{\sqrt{3}}\).
Let \(x_1 = 1 - \frac{1}{\sqrt{3}} \approx 0.42\) and \(x_2 = 1 + \frac{1}{\sqrt{3}} \approx 1.58\).
The parabola \(2+3x^2\) opens upwards. For \(x \in (x_1, x_2)\), the line \(6x\) is above the parabola.
Thus, the maximum transitions between the curves at \(x_1\) and \(x_2\). Since the derivatives of \(6x\) and \(2+3x^2\) are not equal at these points, \(f_1(x)\) is non-differentiable at \(x_1\) and \(x_2\).
Both points lie within the given interval \((-\pi, \pi)\).
Let's analyze \(f_2(x) = |x-1| \cos \left| x^2 - \frac{1}{4} \right|\):
The term \(\cos \left| x^2 - \frac{1}{4} \right|\) simplifies to \(\cos(x^2 - \frac{1}{4})\) because cosine is an even function (\(\cos|A| = \cos A\)).
Since \(x^2 - \frac{1}{4}\) is a smooth polynomial and cosine is a smooth function, the entire term \(\cos(x^2 - \frac{1}{4})\) is differentiable everywhere.
Now look at the term \(|x-1|\). It is non-differentiable at \(x = 1\).
To see if the product is differentiable at \(x=1\), we check the value of the other factor at \(x=1\).
At \(x = 1\), \(\cos(1^2 - \frac{1}{4}) = \cos(\frac{3}{4})\). Since \(\frac{3}{4} \approx 0.75\) radians is not an odd multiple of \(\pi/2\), \(\cos(\frac{3}{4}) \neq 0\).
Therefore, the sharp corner of \(|x-1|\) is not "flattened" by a zero multiplier, meaning \(f_2(x)\) is non-differentiable at \(x = 1\).
Step 4: Final Answer:
The non-differentiable points of \(f_1(x)\) are \(x_1 = 1 - \frac{1}{\sqrt{3}}\) and \(x_2 = 1 + \frac{1}{\sqrt{3}}\).
The non-differentiable point of \(f_2(x)\) is \(x_3 = 1\).
Since all these points are distinct, their sum \(f(x) = f_1(x) + f_2(x)\) will be non-differentiable at exactly these 3 points.