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Step-by-Step Solution
Step 1: Understand the given piecewise definition
The function is given by:
f(x)=
\begin{cases}
|\,4x^2 - 8x + 5\,|, & \text{if } 8x^2 - 6x + 1 \ge 0,\\
\lfloor 4x^2 - 8x + 5 \rfloor, & \text{if } 8x^2 - 6x + 1 < 0,
\end{cases}
where \lfloor \alpha \rfloor denotes the greatest integer less than or equal to \alpha .
Step 2: Determine the interval where 8x^2 - 6x + 1 \ge 0
Solve 8x^2 - 6x + 1 = 0 to find boundaries of the inequality:
8x^2 - 6x + 1 = 0.
The discriminant is \Delta = (-6)^2 - 4 \cdot 8 \cdot 1 = 36 - 32 = 4. Thus, the roots are
x = \frac{6 \pm \sqrt{4}}{16} = \frac{6 \pm 2}{16}.
Hence, the roots are x = \frac{1}{4} and x = \frac{1}{2}. Since 8x^2 - 6x + 1 opens upwards (coefficient of x^2 is positive), the inequality 8x^2 - 6x + 1 \ge 0 holds for
x \in (-\infty,\;\frac{1}{4}] \cup [\frac{1}{2},\;\infty).
In these intervals, f(x) = |\,4x^2 - 8x + 5\,|.
Step 3: Simplify |\,4x^2 - 8x + 5\,| in these intervals
Notice 4x^2 - 8x + 5 = 4(x^2 - 2x) + 5
= 4\bigl(x^2 - 2x + 1\bigr) + 5 - 4
= 4(x - 1)^2 + 1.
This expression 4(x - 1)^2 + 1 is always positive for all real x , so |\,4x^2 - 8x + 5\,| = 4x^2 - 8x + 5.
Therefore, in (-\infty,\;\frac{1}{4}] \cup [\frac{1}{2},\;\infty),
f(x) = 4x^2 - 8x + 5.
Step 4: Determine the interval where 8x^2 - 6x + 1 < 0
From Step 2, we have 8x^2 - 6x + 1 < 0 for x \in \bigl(\frac{1}{4},\;\frac{1}{2}\bigr). In this range,
f(x) = \lfloor 4x^2 - 8x + 5 \rfloor.
Step 5: Examine 4x^2 - 8x + 5 on \left(\frac{1}{4},\,\frac{1}{2}\right)
We still use 4x^2 - 8x + 5 = 4(x-1)^2 + 1. Let us see how its value changes in \left(\frac{1}{4},\;\frac{1}{2}\right).
At x=\frac{1}{4},
4\left(\frac{1}{4}\right)^2 - 8\left(\frac{1}{4}\right) + 5
= 4\left(\frac{1}{16}\right) - 2 + 5
= \frac{1}{4} + 3
= 3.25.
At x=\frac{2-\sqrt{2}}{2}, substitute into 4x^2 - 8x + 5 to check the integer boundary. Through the solution steps (or by direct algebraic simplification) one finds it becomes exactly 3.
At x=\frac{1}{2},
4\left(\frac{1}{2}\right)^2 - 8\left(\frac{1}{2}\right) + 5
= 4\left(\frac{1}{4}\right) - 4 + 5
= 1 - 4 + 5
= 2.
Hence, within \bigl(\frac{1}{4},\,\frac{1}{2}\bigr), the greatest integer function \lfloor 4x^2 - 8x + 5 \rfloor takes different integer values based on these boundaries:
For \frac{1}{4} < x < \frac{2 - \sqrt{2}}{2}, we get values between 3.25 (just over 3) and 3, so the floor is 3.
For \frac{2 - \sqrt{2}}{2} \le x < \frac{1}{2}, the expression is between 3 and 2 , so the floor is 2.
Step 6: Identify potential points of non-differentiability
A piecewise function can fail to be differentiable at points where its definition changes or where the greatest integer function causes discontinuities in the derivative. Here, we check:
x = \frac{1}{4} (transition from 4x^2 - 8x + 5 to \lfloor 4x^2 - 8x + 5 \rfloor )
x = \frac{2-\sqrt{2}}{2} (the value inside the floor changes from 3 to 2 )
x = \frac{1}{2} (transition back to 4x^2 - 8x + 5 from the floor function)
At each of these three points, the expression for f(x) changes form or the floor function jumps to a new integer value, creating a non-differentiable point.
Step 7: Count the number of non-differentiable points
From the above analysis, f(x) is not differentiable at exactly three points:
x = \frac{1}{4}, \quad x = \frac{2-\sqrt{2}}{2}, \quad \text{and} \quad x = \frac{1}{2}.
So, the total number of points of non-differentiability is:
\boxed{3}.
