Let f(x) = { |4x² - 8x + 5|, if 8x² - 6x + 1 ≥ 0; [4x² - 8x + 5], if 8x² - 6x + 1

Q: Why can we just drop the modulus? Isn't it dangerous to assume it's always positive?

That is a fantastic instinct! In mathematics, we never assume; we verify. We looked at the discriminant $D = b^2 - 4ac$ for the inner quadratic $4x^2 - 8x + 5$. Since $D = 64 - 80 = -16$, which is less than zero, and the leading coefficient is positive, the parabola never touches or crosses the x-axis. It stays strictly above it. Therefore, the modulus $|g(x)|$ is identical to $g(x)$ for all real $x$. You aren't being reckless; you are being rigorous!

Q: At $x = 1/2$, the function is continuous. Why is it still non-differentiable?

Continuity is a necessary condition for differentiability, but it is not sufficient. Think of it like this: a road can be continuous (no gaps), but if it has a sharp, jagged turn, you cannot drive smoothly through it. At $x = 1/2$, the left-hand derivative is $0$ (a flat line), but the right-hand derivative is $-4$ (the slope of the parabola). Because the slopes don't match, you have a 'kink' or a 'sharp corner.' The derivative does not exist at that point.

Q: How do I know if I missed a point of non-differentiability?

The best way to be sure is to categorize your search. In piecewise functions, non-differentiability usually hides in three places: 1) The boundary points where the definition changes (like $x=1/4$ and $x=1/2$), 2) Points where the inner function hits an integer (causing the floor function to jump), and 3) Points where the function is continuous but has a 'kink.' By checking these three categories systematically, you ensure no point escapes your analysis.

Question

Let $f(x) = \begin{cases} |4x^2 - 8x + 5|, & 	ext{if } 8x^2 - 6x + 1 \ge 0 \ [4x^2 - 8x + 5], & 	ext{if } 8x^2 - 6x + 1 < 0 \end{cases}$ where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbb{R}$ where $f$ is not differentiable is ______.

Sigma Percentile · Accepted Answer

### The Gatekeeper: Analyzing the Domain

Our function $f(x)$ is defined by the sign of the quadratic $8x^2 - 6x + 1$. This is our gatekeeper.

To understand where the function behaves in which way, we must find the roots of this quadratic. Setting $8x^2 - 6x + 1 = 0$, we factorize it into $(4x - 1)(2x - 1) = 0$.

This gives us two critical points: $x = 1/4$ and $x = 1/2$. Between these roots, the quadratic is negative, triggering the floor function. Outside these roots, it is positive, triggering the modulus. This is the skeleton of our problem.

### The Inner Beauty: Simplifying the Modulus

Now, look at the expression inside both the modulus and the floor function: $g(x) = 4x^2 - 8x + 5$. It looks intimidating, but let us test its nature.

We calculate the discriminant:
$$D = (-8)^2 - 4(4)(5) = 64 - 80 = -16$$

Since the discriminant is negative and the leading coefficient is positive, this parabola never touches the $x$-axis. It is always positive!

This is a moment of pure relief—the modulus sign is essentially doing nothing. We can simply treat it as $g(x) = 4x^2 - 8x + 5$.

### The Staircase: The Floor Function's Jump

In the interval $(1/4, 1/2)$, our function is defined as $[g(x)]$. As $x$ moves from $1/4$ to $1/2$, we calculate the values of $g(x)$.

At $x = 1/4$, $g(1/4) = 3.25$. At $x = 1/2$, $g(1/2) = 2$. So, in this interval, $g(x)$ is sliding down from $3.25$ to $2$.

Because it crosses the integer $3$, the floor function must jump. We solve $4x^2 - 8x + 5 = 3$, which simplifies to:
$$2x^2 - 4x + 1 = 0$$

Using the quadratic formula, we find $x = 1 - 1/\sqrt{2} \approx 0.293$. This is our third critical point.

### The Investigation: Checking Differentiability

Now, we hunt for the points of non-differentiability.

First, at $x = 1/4$: the left-hand limit is $3.25$, but the right-hand limit (the floor of $3.25$) is $3$. This is a **discontinuity**, making it our first point of non-differentiability.

Second, at $x = 1 - 1/\sqrt{2}$: the floor function jumps from $3$ to $2$. Again, this is a **jump discontinuity**, which is our second point.

Finally, at $x = 1/2$: the left-hand limit is $2$ and the right-hand limit is $2$. It is continuous, but we must check the slopes.

The left-hand derivative is $0$, while the right-hand derivative (the slope of $4x^2 - 8x + 5$ at $x=1/2$) is:
$$\frac{d}{dx}(4x^2 - 8x + 5) = 8x - 8 \Rightarrow 8(1/2) - 8 = -4$$

The slopes clash, creating a sharp corner. That is our third point. We have found **three points of non-differentiability**.

Enter Numerical Value:

Visualized Solution

$A na l y z in g t h e P i ece w i se C o n d i t i o n$

$F in d in g t h e C r i t i c a l I n t er v a l s$

$A na l y z in g t h e I nn er Q u a d r a t i c g (x)$

$E v a l u a t in g a tB o u n d a r y P o in t s$

$T h e J u m p in t h e F l oor F u n c t i o n$

$G r a p hin g t h e F l oor F u n c t i o n$

$C h ec k in g D i f f er e n t iabi l i t y a t x = 1/4$

$C h ec k in g D i f f er e n t iabi l i t y a t x = 1 - 1/ 2$

$C h ec k in g D i f f er e n t iabi l i t y a t x = 1/2$

$F ina l C o n c l u s i o n$

The Sigma Insight: Differentiability of a Function

The Gatekeeper

The Inner Beauty

The Staircase

The Investigation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingthePiecewiseCondition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheCriticalIntervals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingtheInnerQuadraticg(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EvaluatingatBoundaryPoints

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheJumpintheFloorFunction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } GraphingtheFloorFunction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CheckingDifferentiabilityatx=1/4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CheckingDifferentiabilityatx=1−1/2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CheckingDifferentiabilityatx=1/2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalConclusion