1. Given piecewise function

We have

where $[\alpha]$ is the greatest integer function.

We must find the number of points in $\mathbb R$ where $f$ is not differentiable.

2. First find where each branch applies

Consider $8x^2-6x+1=(4x-1)(2x-1).$ So, $8x^2-6x+1\ge 0 \iff x\le \frac14 \quad \text{or} \quad x\ge \frac12,$ and $8x^2-6x+1<0 \iff \frac14 < x < \frac12.$

Thus,

• for $x\le \frac14$ or $x\ge \frac12$, $f(x)=|4x^2-8x+5|$,
• for $\frac14<x<\frac12$, $f(x)=[4x^2-8x+5]$.

3. Simplify the quadratic inside

Let $g(x)=4x^2-8x+5=4(x-1)^2+1.$ Since $4(x-1)^2+1\ge 1>0 \quad \text{for all } x,$ we have $|g(x)|=g(x).$

So the outer branch is simply $f(x)=4x^2-8x+5 \quad \text{for } x\le \frac14 \text{ or } x\ge \frac12.$

Hence the function becomes

4. Study the middle interval $\left(\frac14,\frac12\right)$

On this interval, $g(x)=4x^2-8x+5.$ Let us find its range there.

Since $g'(x)=8x-8<0 \quad \text{for all } x<1,$ in particular on $\left(\frac14,\frac12\right)$, $g$ is strictly decreasing.

Now, $g\left(\frac14\right)=4\cdot\frac1{16}-8\cdot\frac14+5=\frac14-2+5=\frac{13}{4}=3.25,$ $g\left(\frac12\right)=4\cdot\frac14-8\cdot\frac12+5=1-4+5=2.$

Since endpoints are not included in the middle interval, $g(x)\in (2,\tfrac{13}{4}) \quad \text{for } \frac14<x<\frac12.$

Therefore,

Since $g(x)<\frac{13}{4}<4$, only values $2$ and $3$ occur.

So the floor-function branch can change only where $g(x)=3.$

5. Find discontinuity/non-differentiability inside the middle interval

Solve $4x^2-8x+5=3.$ This gives $4x^2-8x+2=0$ $2x^2-4x+1=0$ $x=\frac{4\pm\sqrt{16-8}}{4}=1\pm \frac{\sqrt2}{2}.$

Among these, only $x=1-\frac{\sqrt2}{2}$ lies in $\left(\frac14,\frac12\right)$.

At this point, the floor function jumps from $3$ to $2$, so $f$ is discontinuous there, hence not differentiable there.

So one non-differentiable point is $x=1-\frac{\sqrt2}{2}.$

6. Check the boundary points $x=\frac14$ and $x=\frac12$

These are where the definition changes, so they must be tested carefully.

At $x=\frac14$

Since $x=\frac14$ belongs to the first branch, $f\left(\frac14\right)=g\left(\frac14\right)=\frac{13}{4}.$

For $x\to \left(\frac14\right)^-$, $f(x)=g(x) \to \frac{13}{4}.$

For $x\to \left(\frac14\right)^+$, we are in the floor branch. Since on $\left(\frac14,\frac12\right)$, $g(x)<\frac{13}{4}<4 \quad \text{and} \quad g(x)>3,$ so $[g(x)]=3.$ Thus $\lim_{x\to (1/4)^+} f(x)=3 \ne \frac{13}{4}.$

Hence $f$ is discontinuous at $x=\frac14$, so it is not differentiable there.

At $x=\frac12$

Since $x=\frac12$ belongs to the first branch, $f\left(\frac12\right)=g\left(\frac12\right)=2.$

For $x\to \left(\frac12\right)^-$, in the floor branch we have $g(x)>2$ but close to $2$, so $[g(x)]=2.$ Thus $\lim_{x\to (1/2)^-} f(x)=2.$

For $x\to \left(\frac12\right)^+$, $f(x)=g(x) \to 2.$ So $f$ is continuous at $x=\frac12$.

Now check differentiability:

• Left of $\frac12$, $f(x)=2$ in a neighborhood sufficiently close to $\frac12$ (inside the middle interval), so $f'_-(\tfrac12)=0.$
• Right of $\frac12$, $f(x)=g(x)$, so $f'_+(\tfrac12)=g'(\tfrac12)=8\cdot\frac12-8=-4.$

Since $f'_-(\tfrac12) \ne f'_+(\tfrac12),$ $f$ is not differentiable at $x=\frac12$.

7. Count all non-differentiable points

They are:

1. $x=\frac14$
2. $x=1-\frac{\sqrt2}{2}$
3. $x=\frac12$

So the total number of points is $\boxed{3}.$

8. Comparison with stored answer

Derived answer = $3$. Stored correct answer = $3$.

They agree.