To determine the points at which the function \( f(x) = \max\{6x, 2 + 3x^2\} + |x - 1| \cos|x^2 - \frac{1}{4}| \) is not differentiable, we need to consider the differentiability of both components: the max function and the absolute value function.
The function \( g(x) = \max\{6x, 2 + 3x^2\} \) is not differentiable where the expressions inside the max are equal, i.e., \( 6x = 2 + 3x^2 \). This simplifies to:
\( 3x^2 - 6x + 2 = 0 \).
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 3, b = -6, c = 2 \), gives:
\( x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3} \).
These are the points where \( g(x) \) is not differentiable.
Next, consider \( h(x) = |x - 1| \cos|x^2 - \frac{1}{4}| \), which is not differentiable where \( x - 1 = 0 \), i.e., \( x = 1 \).
Now, combine these potential non-differentiable points:
| \(1 + \frac{\sqrt{3}}{3}\) |
| \(1 - \frac{\sqrt{3}}{3}\) |
| 1 |
All points lie within \( x \in (-\pi, \pi) \).
Thus, the number of points where \( f(x) \) is not differentiable is 3.
The result, 3, verifies the given range 3,3.
Let $f(x) = \begin{cases} e^{x-1} & , x < 0 \\ x^2 - 5x + 6 & , x \geq 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to ________