Analyzing Function F(x): Domain And Continuity

Hey guys! Let's dive into analyzing a piecewise function, figuring out its domain, and checking if it's continuous. We'll break it down step by step so it's super clear.

Defining the Function f(x)

First, let's clearly define the function we're working with. The function $f(x)$ is defined piecewise as follows:

$f(x)=\left\{\begin{array}{ll} 2^x, & x<0 \\ -x^2-4 x+1, & 0 \\ \frac{1}{2} x+3, & x>2 \end{array}\right.$

This means that for different intervals of $x$ values, we use different formulas to calculate the value of $f(x)$. Understanding this is crucial before we can analyze its properties.

Domain of the Function

Understanding the Domain: The domain of a function is the set of all possible input values (x-values) for which the function is defined. In other words, it's all the $x$ values you can plug into the function and get a valid output.

For our piecewise function, we need to check the domain for each piece:

1. $f(x) = 2^x$ for $x < 0$: The exponential function $2^x$ is defined for all real numbers. So, for $x < 0$, there are no restrictions.
2. $f(x) = -x^2 - 4x + 1$ for $0 \le x \le 2$: This is a quadratic function, and quadratic functions are defined for all real numbers. Therefore, in the interval $0 \le x \le 2$, there are no restrictions.
3. $f(x) = \frac{1}{2}x + 3$ for $x > 2$: This is a linear function, which is also defined for all real numbers. So, for $x > 2$, there are no restrictions.

Combining the Intervals: Now, let's look at the intervals themselves:

• $x < 0$
• $0 \le x \le 2$
• $x > 2$

Notice that these intervals cover all real numbers. There are no gaps or undefined points. The first interval covers all numbers less than 0, the second interval covers numbers from 0 to 2 (inclusive), and the third interval covers all numbers greater than 2. Together, they include every single real number.

Conclusion on Domain: Since each piece of the function is defined on its respective interval, and the intervals together cover all real numbers, the domain of the entire function $f(x)$ is all real numbers. This means you can input any real number into this function, and it will give you a valid output.

So, the statement "The domain is all real numbers" is TRUE for the function $f(x)$.

Continuity of the Function

What is Continuity?: A function is continuous if you can draw its graph without lifting your pen. More formally, a function $f(x)$ is continuous at a point $x = a$ if the following three conditions are met:

1. $f(a)$ is defined (the function has a value at $x = a$).
2. $\lim_{x \to a} f(x)$ exists (the limit of the function as $x$ approaches $a$ exists).
3. $\lim_{x \to a} f(x) = f(a)$ (the limit of the function as $x$ approaches $a$ is equal to the function's value at $x = a$).

For a piecewise function, we need to check continuity at the points where the function definition changes. These are the points where the pieces "meet." In our case, we need to check continuity at $x = 0$ and $x = 2$.

Continuity at x = 0

1. Check if f(0) is defined: According to the function definition, $f(x) = -x^2 - 4x + 1$ for $0 \le x \le 2$. So, $f(0) = -(0)^2 - 4(0) + 1 = 1$. Thus, $f(0)$ is defined and equals 1.

2. Check if the limit exists: We need to check the left-hand limit and the right-hand limit as $x$ approaches 0.

   • Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2^x = 2^0 = 1$
   • Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-x^2 - 4x + 1) = -(0)^2 - 4(0) + 1 = 1$

   Since the left-hand limit and the right-hand limit are equal, the limit exists, and $\lim_{x \to 0} f(x) = 1$.

3. Check if the limit equals the function value: We have $\lim_{x \to 0} f(x) = 1$ and $f(0) = 1$. Therefore, $\lim_{x \to 0} f(x) = f(0)$.

Since all three conditions are met at $x = 0$, the function is continuous at $x = 0$.

Continuity at x = 2

1. Check if f(2) is defined: According to the function definition, $f(x) = -x^2 - 4x + 1$ for $0 \le x \le 2$. So, $f(2) = -(2)^2 - 4(2) + 1 = -4 - 8 + 1 = -11$. Thus, $f(2)$ is defined and equals -11.

2. Check if the limit exists: We need to check the left-hand limit and the right-hand limit as $x$ approaches 2.

   • Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (-x^2 - 4x + 1) = -(2)^2 - 4(2) + 1 = -4 - 8 + 1 = -11$
   • Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (\frac{1}{2}x + 3) = \frac{1}{2}(2) + 3 = 1 + 3 = 4$

   Since the left-hand limit (-11) and the right-hand limit (4) are not equal, the limit does not exist at $x = 2$.

3. Conclusion: Because the limit does not exist at $x = 2$, the function is not continuous at $x = 2$.

Overall Continuity: Since the function is not continuous at $x = 2$, the function $f(x)$ is not continuous everywhere. It has a discontinuity at $x = 2$.

Therefore, the statement "The function is continuous" is FALSE for the function $f(x)$.

Final Answer

In summary:

• The domain of the function $f(x)$ is all real numbers.
• The function $f(x)$ is not continuous because it is discontinuous at $x = 2$.

So, out of the given statements, only the statement about the domain being all real numbers is true. Hope this helps you guys understand how to analyze piecewise functions!