Function Analysis: Expression, Continuity, And Strictness

Hey guys! Let's dive into an exciting math problem that involves analyzing a function, determining its expression over specific intervals, and checking its continuity and strictness. This should be fun!

Understanding the Function

Before we begin, let's make sure we understand what the function $f(x)$ is all about. We are given that $f(x) = E(x) + (x - E(x))^2$, where $E(x)$ represents the floor function, also known as the greatest integer function. In simple terms, $E(x)$ gives you the greatest integer less than or equal to $x$. For example, $E(3.14) = 3$, $E(5) = 5$, and $E(-2.7) = -3$.

Determining the Expression of $f(x)$ on Intervals $[k, k+1[$

Okay, so the first task is to figure out the expression for $f(x)$ on each interval $[k, k+1[$ for any integer $k$. This means we are looking at intervals like $[0, 1[$, $[1, 2[$, $[-1, 0[$, and so on. The key here is to recognize that within each of these intervals, the value of $E(x)$ is constant. Specifically, for any $x$ in the interval $[k, k+1[$, we have $E(x) = k$.

Now, let's substitute $E(x) = k$ into the function's definition:

$f(x) = E(x) + (x - E(x))^2 = k + (x - k)^2$

So, on the interval $[k, k+1[$, the function $f(x)$ can be expressed as $f(x) = k + (x - k)^2$. This is a quadratic function within each interval, shifted vertically by $k$.

For example:

• On the interval $[0, 1[$, $f(x) = 0 + (x - 0)^2 = x^2$
• On the interval $[1, 2[$, $f(x) = 1 + (x - 1)^2$
• On the interval $[-1, 0[$, $f(x) = -1 + (x + 1)^2$

In summary, for any integer $k$, the expression of $f(x)$ on the interval $[k, k+1[$ is given by $f(x) = k + (x - k)^2$. This is a crucial result as it simplifies the function's behavior within each interval, making it easier to analyze its continuity and strictness. Keep in mind that understanding the floor function is key to tackling problems like these. The floor function, denoted as $E(x)$, returns the greatest integer less than or equal to $x$. In mathematical notation, $E(x) = \{ n \in \mathbb{Z} \mid n \leq x < n+1 \}$. For instance, $E(3.14) = 3$, $E(-2.5) = -3$, and $E(5) = 5$. The floor function is piecewise constant, meaning its value remains constant over intervals of the form $[k, k+1)$ where $k$ is an integer. Understanding this property is crucial for analyzing functions that involve the floor function, such as the one in this problem. Specifically, within each interval $[k, k+1)$, $E(x) = k$, which simplifies the analysis significantly.

Proving Continuity and Strictness

Alright, now let's tackle the second part of the problem: showing whether $f(x)$ is continuous and strictly increasing. This involves a bit more work but is definitely manageable.

Continuity of $f(x)$

To prove continuity, we need to show that $f(x)$ is continuous at every point in its domain, which is $\mathbb{R}$. Since $f(x)$ is defined piecewise on intervals $[k, k+1[$, we need to check continuity at the integer points $x = k$ for all $k \in \mathbb{Z}$. In other words, we want to ensure that the function smoothly transitions from one interval to the next without any jumps or breaks. For $f(x)$ to be continuous at $x = k$, the left-hand limit and the right-hand limit at $x = k$ must exist and be equal to the function's value at $x = k$. Mathematically, this means:

$\lim_{x \to k^-} f(x) = \lim_{x \to k^+} f(x) = f(k)$

Let's analyze the left-hand limit. As $x$ approaches $k$ from the left, $x$ lies in the interval $[k-1, k[$. Thus, according to our earlier result:

$f(x) = k - 1 + (x - (k - 1))^2$

So, the left-hand limit is:

$\lim_{x \to k^-} f(x) = \lim_{x \to k^-} [k - 1 + (x - k + 1)^2] = k - 1 + (k - k + 1)^2 = k - 1 + 1 = k$

Now, let's analyze the right-hand limit. As $x$ approaches $k$ from the right, $x$ lies in the interval $[k, k+1[$. Thus, according to our earlier result:

$f(x) = k + (x - k)^2$

So, the right-hand limit is:

$\lim_{x \to k^+} f(x) = \lim_{x \to k^+} [k + (x - k)^2] = k + (k - k)^2 = k + 0 = k$

Finally, let's find the value of $f(x)$ at $x = k$. Since $E(k) = k$, we have:

$f(k) = E(k) + (k - E(k))^2 = k + (k - k)^2 = k + 0 = k$

Since $\lim_{x \to k^-} f(x) = \lim_{x \to k^+} f(x) = f(k) = k$, we can conclude that $f(x)$ is continuous at $x = k$ for all integers $k$. Because $f(x)$ is also continuous within each interval $[k, k+1[$ (as it is a quadratic function), we can confidently say that $f(x)$ is continuous on $\mathbb{R}$. Understanding the concept of limits is crucial for proving continuity. The limit of a function at a point is the value that the function approaches as the input approaches that point. In the context of continuity, we need to ensure that the left-hand limit, right-hand limit, and the function value at a point are all equal. If they are, then the function is continuous at that point; otherwise, it is discontinuous. This is precisely what we demonstrated when we analyzed the continuity of $f(x)$ at integer points.

Strictness of $f(x)$

Next, we need to determine if $f(x)$ is strictly increasing. A function is strictly increasing if, for any $x_1$ and $x_2$ in its domain, $x_1 < x_2$ implies $f(x_1) < f(x_2)$. To investigate this, let's consider two cases:

Case 1: $x_1$ and $x_2$ are in the same interval $[k, k+1[$

In this case, $f(x) = k + (x - k)^2$. The derivative of $f(x)$ with respect to $x$ is:

$f'(x) = 2(x - k)$

Since $x > k$ in the interval $[k, k+1[$, $f'(x) > 0$. This means that $f(x)$ is strictly increasing within each interval $[k, k+1[$.

Case 2: $x_1$ is in the interval $[k, k+1[$ and $x_2$ is in the interval $[k+1, k+2[$

Here, we have $x_1 < k+1 \leq x_2$. We need to show that $f(x_1) < f(x_2)$. We know that:

$f(x_1) = k + (x_1 - k)^2$

and

$f(x_2) = k + 1 + (x_2 - (k + 1))^2$

To compare $f(x_1)$ and $f(x_2)$, we need to consider the maximum possible value of $f(x_1)$ and the minimum possible value of $f(x_2)$. The maximum value of $f(x_1)$ occurs as $x_1$ approaches $k+1$ from the left, so:

$\lim_{x_1 \to (k+1)^-} f(x_1) = k + ((k+1) - k)^2 = k + 1$

The minimum value of $f(x_2)$ occurs when $x_2 = k+1$, so:

$f(k+1) = k + 1 + ((k+1) - (k + 1))^2 = k + 1$

This shows that it is not always true that $f(x_1) < f(x_2)$. For instance, let's take $x_1$ very close to 1 from the left (say, 0.99) and $x_2 = 1$. Then:

$f(0.99) = 0 + (0.99 - 0)^2 = 0.9801$

$f(1) = 1 + (1 - 1)^2 = 1$

In this case, $f(0.99) < f(1)$. However, if we consider $x_1 = 0$ and $x_2 = 1$, we have:

$f(0) = 0 + (0 - 0)^2 = 0$

$f(1) = 1 + (1 - 1)^2 = 1$

Therefore, $f(x)$ is not strictly increasing on $\mathbb{R}$. Although $f(x)$ is strictly increasing within each interval $[k, k+1[$, it is not strictly increasing across the entire real line due to the jumps at integer values. Analyzing the derivative of the function can provide valuable insights into its increasing or decreasing behavior. If the derivative is positive over an interval, the function is increasing; if it is negative, the function is decreasing. However, when dealing with piecewise functions or functions with discontinuities, it is essential to consider the behavior at the points where the function definition changes or where discontinuities occur, as these points may affect the overall increasing or decreasing nature of the function. This analysis confirms that while $f(x)$ is strictly increasing within each interval $[k, k+1)$, it is not strictly increasing on $\mathbb{R}$.

Conclusion

So, to wrap it up:

• The expression of $f(x)$ on the interval $[k, k+1[$ is $f(x) = k + (x - k)^2$.
• $f(x)$ is continuous on $\mathbb{R}$.
• $f(x)$ is not strictly increasing on $\mathbb{R}$.

Hope that helped clear things up! Keep practicing and you'll become a math whiz in no time! Analyzing functions can be challenging, but by breaking down the problem into smaller, manageable steps, you can tackle even the most complex functions. Remember to pay close attention to the function's definition, its domain, and any special properties it may have, such as discontinuities or piecewise behavior. By combining your understanding of these concepts with techniques like finding limits and derivatives, you'll be well-equipped to analyze a wide range of functions and uncover their hidden behaviors.