Let f : (0,1) → ℝ be the function defined as f(x) = [4x] (x - 1/4)² (x - 1/2), where [x] denotes the

Q: Wait, why do we only check $x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$? What about other points?

Great question! The Greatest Integer Function $[4x]$ is constant between integers. It only 'jumps' or changes its definition when the input $4x$ hits an integer. Everywhere else, the function is just a simple polynomial, which is inherently continuous and differentiable. The 'action' only happens at the boundaries.

Q: Why is the function continuous at $x = \frac{1}{2}$ but not differentiable?

Continuity just means the graph doesn't break—the left and right sides meet at the same point. Differentiability is stricter; it requires the *slope* (the tangent) to be the same from both sides. At $x = \frac{1}{2}$, the graph connects smoothly, but the steepness changes abruptly because of the multiplier, creating a 'kink' in the graph.

Q: How do I know where the minimum value is without checking every single point?

We use the sign of the function! In the interval $(0, \frac{1}{4})$, the function is $0$. In $(\frac{1}{4}, \frac{1}{2})$, the term $(x - \frac{1}{2})$ is negative, making the whole function negative. After $\frac{1}{2}$, everything is positive. Thus, the minimum *must* be in the second interval. We only need to find the local minimum of that specific piece.

Question

Let $f : (0,1) 	o \mathbb{R}$ be the function defined as $f(x) = [4x] \left(x - \frac{1}{4}ight)^2 \left(x - \frac{1}{2}ight)$, where $[x]$ denotes the greatest integer less than or equal to $x$. Then which of the following statements is(are) true?

Options:

(A) The function $f$ is discontinuous exactly at one point in $(0,1)$

(B) There is exactly one point in $(0,1)$ at which the function $f$ is continuous but NOT differentiable

(C) The function $f$ is NOT differentiable at more than three points in $(0,1)$

(D) The minimum value of the function $f$ is $-\frac{1}{512}$

Sigma Percentile · Accepted Answer

### The Symphony of Piecewise Functions: Decoding the Greatest Integer Trap

My dear student, welcome to a problem that tests not just your calculation skills, but your ability to visualize the behavior of functions. We are looking at $f(x) = [4x] (x - \frac{1}{4})^2 (x - \frac{1}{2})$.

At first glance, it looks like a standard polynomial, but that $[4x]$—the Greatest Integer Function—is a wolf in sheep's clothing. It is the heartbeat of this problem, dictating when the function changes its personality. Let us peel back the layers together.

### Phase 1: The Anatomy of the Function

The Greatest Integer Function $[4x]$ is a step function. It remains constant until $4x$ hits an integer, at which point it jumps to the next value.

In the interval $(0, 1)$, $4x$ becomes an integer exactly when $x = \frac{1}{4}, \frac{1}{2}, 	ext{and } \frac{3}{4}$. These are our critical junction points. We define our function piecewise as follows:

For $x \in (0, \frac{1}{4})$, $[4x] = 0$, so $f(x) = 0$.

For $x \in [\frac{1}{4}, \frac{1}{2})$, $[4x] = 1$, so $f(x) = (x - \frac{1}{4})^2 (x - \frac{1}{2})$.

For $x \in [\frac{1}{2}, \frac{3}{4})$, $[4x] = 2$, so $f(x) = 2(x - \frac{1}{4})^2 (x - \frac{1}{2})$.

For $x \in [\frac{3}{4}, 1)$, $[4x] = 3$, so $f(x) = 3(x - \frac{1}{4})^2 (x - \frac{1}{2})$.

### Phase 2: The Continuity Check

Continuity is about connection. Does the graph break?

At $x = \frac{1}{4}$, the Left-Hand Limit (LHL) is $0$. The Right-Hand Limit (RHL) is $(x - \frac{1}{4})^2 (x - \frac{1}{2})$, which at $x = \frac{1}{4}$ is also $0$. They meet, so it is continuous.

At $x = \frac{1}{2}$, the LHL is $(x - \frac{1}{4})^2 (x - \frac{1}{2}) = 0$, and the RHL is $2(x - \frac{1}{4})^2 (x - \frac{1}{2}) = 0$. Again, they meet, so it is continuous.

At $x = \frac{3}{4}$, the LHL is $2(\frac{1}{2})^2(\frac{1}{4}) = \frac{1}{8}$, while the RHL is $3(\frac{1}{2})^2(\frac{1}{4}) = \frac{3}{16}$. They do not meet, meaning the graph breaks. Thus, the function is **discontinuous at $x = \frac{3}{4}$**.

### Phase 3: The Differentiability Dance

A function is differentiable only if the slope from the left matches the slope from the right. Let $g(x) = (x - \frac{1}{4})^2 (x - \frac{1}{2})$.

Using the product rule, the derivative is:
$$g'(x) = (x - \frac{1}{4})(3x - \frac{5}{4})$$

At $x = \frac{1}{4}$, the Left-Hand Derivative (LHD) is $0$ (since $f(x)=0$ on the left), and the Right-Hand Derivative (RHD) is $g'(\frac{1}{4}) = 0$. They match, so it is differentiable.

At $x = \frac{1}{2}$, the LHD is $g'(\frac{1}{2}) = \frac{1}{16}$, but the RHD is $2 \cdot g'(\frac{1}{2}) = \frac{1}{8}$. They do not match. The function has a 'kink' here; it is continuous but **not differentiable**.

### Phase 4: The Hunt for the Minimum

We seek the minimum value. We know the function is $0$ in the first interval, positive in the third and fourth, and negative in the second interval $([\frac{1}{4}, \frac{1}{2}))$.

Therefore, the global minimum must reside in this second interval. We set the derivative $g'(x) = 0$, which yields $x = \frac{1}{4}$ and $x = \frac{5}{12}$.

Evaluating $f(\frac{5}{12})$ yields:
$$f\left(\frac{5}{12}ight) = \left(\frac{5}{12} - \frac{3}{12}ight)^2 \left(\frac{5}{12} - \frac{6}{12}ight) = \left(\frac{2}{12}ight)^2 \left(-\frac{1}{12}ight) = \frac{1}{36} \cdot \left(-\frac{1}{12}ight) = -\frac{1}{432}$$

Since $-\frac{1}{432} 
eq -\frac{1}{512}$, the proposed minimum value is false. We have navigated the complexities, analyzed the jumps, and found the truth. Keep this rigor in your heart, and no problem will ever be too daunting.

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Visualized Solution

$U n d er s t an d in g [4 x]$

$P i ece w i seD e f ini t i o n o f f (x)$

$C o n t in u i t y a t x = \frac{1}{4}$

$C o n t in u i t y a t x = \frac{1}{2}$

$C o n t in u i t y a t x = \frac{3}{4}$

$D er i v a t i v eo f t h e B a se F u n c t i o n$

$D i f f er e n t iabi l i t y a t x = \frac{1}{4}$

$D i f f er e n t iabi l i t y a t x = \frac{1}{2}$

$E v a l u a t in g O pt i o n s B an d C$

$F in d in g t h e M inim u mV a l u e$

$C a l c u l a t in g t h e M inim u m$

The Sigma Insight: Differentiability of a Function

The Symphony of Piecewise Functions

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

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