Let [t] denote the greatest integer less than or equal to t. Let f : [0, ∞) → R be a function define

Q: Wait, why do we check $x=4$ separately? Shouldn't it be discontinuous because both $g(x)$ and $h(x)$ are?

That is a brilliant question! It is a common trap. While it is true that both $g(x)$ and $h(x)$ are discontinuous at $x=4$, continuity is about the limit of the *entire* function $f(x)$. If the jump in $g(x)$ is exactly equal to the jump in $h(x)$, they can cancel each other out, making $f(x)$ continuous. Always check the limits $LHL$ and $RHL$ for the combined function!

Q: Is $x=0$ a point of discontinuity?

In this problem, we are looking at the interval $[0, 8]$. At $x=0$, the function is defined as $f(0) = [0+3] - [0] = 3$. Since we are only considering the interval $[0, 8]$, we only care about the right-hand limit at $x=0$. As $x \to 0^+$, the function approaches $f(0)$, so it is continuous from the right at the boundary. Therefore, we do not include $x=0$ in our set $S$.

Q: Why can't we just use the property that the sum or difference of two discontinuous functions is always discontinuous?

That property is actually a misconception! The sum or difference of two discontinuous functions *can* be continuous if the discontinuities 'cancel out'. Think of it like adding two waves; if one has a peak where the other has a trough, they can flatten out. Always verify by checking the limits of the combined function.

Question

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f : [0, \infty) 	o \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a grand, infinite staircase. Each step represents an integer value, and the Greatest Integer Function, denoted by $[t]$, is the mathematical embodiment of this staircase.

It is a function that remains flat, holding its value steady, until it suddenly leaps to the next level the moment its input hits an integer. Today, we are going to explore the function:

$$f(x) = \left[ \frac{x}{2} + 3 ight] - [\sqrt{x}]$$

Our mission is to find the points on the interval $[0, 8]$ where this function, a difference of two such staircases, loses its footing and breaks continuity.

### Phase 1: Analyzing the First Staircase

Let us first isolate the component $g(x) = [\frac{x}{2} + 3]$. The greatest integer function $[t]$ is discontinuous precisely when $t$ is an integer.

Therefore, $g(x)$ will experience a jump whenever the argument $\frac{x}{2} + 3$ becomes an integer. This happens when $\frac{x}{2}$ is an integer, which implies $x$ must be an even number.

Within our interval $[0, 8]$, the points where $g(x)$ jumps are $x \in \{2, 4, 6, 8\}$. At these points, the function $g(x)$ is essentially 'climbing' to a new level.

### Phase 2: Analyzing the Second Staircase

Now, let us turn our attention to the second component, $h(x) = [\sqrt{x}]$. Using the same logic, this function will have a discontinuity whenever $\sqrt{x}$ is an integer.

Squaring both sides, we find that $x$ must be a perfect square. Looking at our interval $[0, 8]$, the perfect squares are $1$ and $4$.

Thus, $h(x)$ jumps at $x=1$ and $x=4$.

### Phase 3: The Great Cancellation Trap

We now have a list of candidate points where $f(x)$ might be discontinuous: $\{1, 2, 4, 6, 8\}$. A fundamental theorem of limits tells us that $f(x)$ can only be discontinuous where at least one of its components is discontinuous.

However, we must be vigilant. Look at $x=4$, where both $g(x)$ and $h(x)$ are discontinuous. Does this mean $f(x)$ is definitely discontinuous? Not necessarily!

Let us calculate the value of $f(4)$:

$$f(4) = \left[ \frac{4}{2} + 3 ight] - [\sqrt{4}] = [5] - [2] = 3$$

Now, let us examine the Left-Hand Limit ($LHL$) as $x 	o 4^-$. For $x$ slightly less than $4$, say $3.9$:

$$g(3.9) = [1.95 + 3] = [4.95] = 4$$
$$h(3.9) = [\sqrt{3.9}] \approx [1.97] = 1$$
$$LHL = 4 - 1 = 3$$

Now, examine the Right-Hand Limit ($RHL$) as $x 	o 4^+$. For $x$ slightly more than $4$, say $4.1$:

$$g(4.1) = [2.05 + 3] = [5.05] = 5$$
$$h(4.1) = [\sqrt{4.1}] \approx [2.02] = 2$$
$$RHL = 5 - 2 = 3$$

Incredibly, $LHL = RHL = f(4) = 3$. The jumps in $g(x)$ and $h(x)$ have perfectly canceled each other out, meaning the function $f(x)$ is **continuous** at $x=4$.

### Phase 4: The Final Summation

Having cleared the trap at $x=4$, we are left with the remaining points: $x=1, 2, 6, 8$. At these points, only one of the two functions jumps, meaning there is no cancellation.

Thus, $f(x)$ is indeed discontinuous at these locations. The set of points of discontinuity is $S = \{1, 2, 6, 8\}$.

The final step of our journey is to sum these values:

$$\sum_{a \in S} a = 1 + 2 + 6 + 8 = 17$$

We have successfully navigated the staircases and found our answer. The final result is **17**.

Enter Numerical Value:

Visualized Solution

Introduction to the Function $f (x)$

Analyzing the First Term $g (x)$

Discontinuities of $g (x)$

Analyzing the Second Term $h (x)$

Discontinuities of $h (x)$

Candidate Points for $f (x)$

Checking Continuity at $x = 4$

Right Hand Limit at $x = 4$

Checking $x = 1$ and $x = 2$

Checking $x = 6$ and $x = 8$

Final Summation

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

Checking Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4

Right Hand Limit at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4

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

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=6 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=8