10. Given The Function $f(x) = -2 \cdot 2^{\frac{1}{3} X} + 3$, Answer The Following:- Domain: - Range:- $y$-intercept:- Asymptote:- Increasing Interval:- Decreasing Interval:- End Behavior:

The given function is $f(x) = -2 \cdot 2^{\frac{1}{3} x} + 3$. To analyze its properties, we need to understand the behavior of the function as $x$ varies. In this article, we will explore the domain, range, $y$-intercept, asymptote, increasing and decreasing intervals, and end behavior of the function.

Domain

The domain of a function is the set of all possible input values for which the function is defined. In this case, the function is defined for all real numbers, as there are no restrictions on the input value $x$. Therefore, the domain of the function is:

• Domain: $(-\infty, \infty)$

Range

The range of a function is the set of all possible output values. To determine the range, we need to find the minimum and maximum values of the function. Since the function is a continuous and differentiable function, we can use calculus to find the critical points and determine the range.

The function can be rewritten as $f(x) = -2 \cdot (2^{\frac{1}{3} x}) + 3$. Let $y = 2^{\frac{1}{3} x}$. Then, $f(x) = -2y + 3$. The minimum value of $f(x)$ occurs when $y$ is maximum, and the maximum value of $f(x)$ occurs when $y$ is minimum.

To find the maximum and minimum values of $y$, we can take the derivative of $y$ with respect to $x$ and set it equal to zero:

$$\frac{dy}{dx} = \frac{1}{3} \cdot 2^{\frac{1}{3} x} \cdot \ln(2) = 0$$

Solving for $x$, we get:

$$x = -\infty$$

Since $x$ cannot be negative infinity, we conclude that the function has no maximum or minimum value. Therefore, the range of the function is:

• Range: $(-\infty, \infty)$

$y$-intercept

The $y$-intercept of a function is the value of the function when $x$ is equal to zero. To find the $y$-intercept, we substitute $x = 0$ into the function:

$$f(0) = -2 \cdot 2^{\frac{1}{3} \cdot 0} + 3 = -2 + 3 = 1$$

Therefore, the $y$-intercept of the function is:

• $y$-intercept: $1$

Asymptote

An asymptote is a line that the function approaches as $x$ goes to positive or negative infinity. To determine the asymptote, we can analyze the behavior of the function as $x$ goes to positive or negative infinity.

As $x$ goes to positive infinity, the term $2^{\frac{1}{3} x}$ grows exponentially, and the function approaches negative infinity. Therefore, the function has a horizontal asymptote at $y = -\infty$.

As $x$ goes to negative infinity, the term $2^{\frac{1}{3} x}$ approaches zero, and the function approaches $3$. Therefore, the function has a horizontal asymptote at $y = 3$.

Therefore, the asymptote of the function is:

• Asymptote: $y = -\infty$ and $y = 3$

Increasing Interval

The increasing interval of a function is the interval where the function is increasing. To determine the increasing interval, we can analyze the behavior of the function's derivative.

The derivative of the function is:

$$f'(x) = -2 \cdot \frac{1}{3} \cdot 2^{\frac{1}{3} x} \cdot \ln(2)$$

The function is increasing when the derivative is positive. Therefore, we need to find the values of $x$ for which the derivative is positive.

Simplifying the derivative, we get:

$$f'(x) = -\frac{2}{3} \cdot 2^{\frac{1}{3} x} \cdot \ln(2)$$

The derivative is positive when $2^{\frac{1}{3} x}$ is negative. This occurs when $x$ is negative.

Therefore, the increasing interval of the function is:

• Increasing Interval: $(-\infty, 0)$

Decreasing Interval

The decreasing interval of a function is the interval where the function is decreasing. To determine the decreasing interval, we can analyze the behavior of the function's derivative.

The derivative of the function is:

$$f'(x) = -2 \cdot \frac{1}{3} \cdot 2^{\frac{1}{3} x} \cdot \ln(2)$$

The function is decreasing when the derivative is negative. Therefore, we need to find the values of $x$ for which the derivative is negative.

Simplifying the derivative, we get:

$$f'(x) = -\frac{2}{3} \cdot 2^{\frac{1}{3} x} \cdot \ln(2)$$

The derivative is negative when $2^{\frac{1}{3} x}$ is positive. This occurs when $x$ is positive.

Therefore, the decreasing interval of the function is:

• Decreasing Interval: $(0, \infty)$

End Behavior

The end behavior of a function is the behavior of the function as $x$ goes to positive or negative infinity. To determine the end behavior, we can analyze the behavior of the function's asymptotes.

As $x$ goes to positive infinity, the function approaches negative infinity. Therefore, the end behavior of the function is:

• End Behavior: $y = -\infty$ as $x \to \infty$

As $x$ goes to negative infinity, the function approaches $3$. Therefore, the end behavior of the function is:

• End Behavior: $y = 3$ as $x \to -\infty$

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In the previous article, we explored the properties of the function $f(x) = -2 \cdot 2^{\frac{1}{3} x} + 3$. In this article, we will answer some frequently asked questions about the function and its properties.

Q: What is the domain of the function?

A: The domain of the function is $(-\infty, \infty)$, which means that the function is defined for all real numbers.

Q: What is the range of the function?

A: The range of the function is also $(-\infty, \infty)$, which means that the function can take on any real value.

Q: What is the $y$-intercept of the function?

A: The $y$-intercept of the function is $1$, which means that when $x = 0$, the function evaluates to $1$.

Q: What is the horizontal asymptote of the function?

A: The horizontal asymptote of the function is $y = -\infty$ as $x \to \infty$ and $y = 3$ as $x \to -\infty$.

Q: What is the increasing interval of the function?

A: The increasing interval of the function is $(-\infty, 0)$, which means that the function is increasing for all values of $x$ less than $0$.

Q: What is the decreasing interval of the function?

A: The decreasing interval of the function is $(0, \infty)$, which means that the function is decreasing for all values of $x$ greater than $0$.

Q: What is the end behavior of the function?

A: The end behavior of the function is $y = -\infty$ as $x \to \infty$ and $y = 3$ as $x \to -\infty$.

Q: How do I graph the function?

A: To graph the function, you can use a graphing calculator or software. You can also use the properties of the function to sketch the graph. For example, you can use the $y$-intercept to find the point where the graph intersects the $y$-axis, and you can use the horizontal asymptotes to find the points where the graph approaches the $x$-axis.

Q: How do I find the derivative of the function?

A: To find the derivative of the function, you can use the power rule and the chain rule. The derivative of the function is $f'(x) = -\frac{2}{3} \cdot 2^{\frac{1}{3} x} \cdot \ln(2)$.

Q: How do I find the second derivative of the function?

A: To find the second derivative of the function, you can differentiate the first derivative. The second derivative of the function is $f''(x) = -\frac{2}{9} \cdot 2^{\frac{1}{3} x} \cdot \ln^2(2)$.

Q: How do I use the function in real-world applications?

A: The function $f(x) = -2 \cdot 2^{\frac{1}{3} x} + 3$ can be used in a variety of real-world applications, such as modeling population growth, chemical reactions, and electrical circuits. The function can also be used to model the behavior of complex systems, such as financial markets and social networks.

Q: How do I extend the function to more complex models?

A: To extend the function to more complex models, you can use a variety of techniques, such as adding more terms to the function, using different types of functions, and incorporating additional variables. You can also use numerical methods, such as the Runge-Kutta method, to solve the differential equations that arise from the function.

In conclusion, the function $f(x) = -2 \cdot 2^{\frac{1}{3} x} + 3$ is a powerful tool for modeling complex systems and phenomena. By understanding the properties of the function and its derivatives, you can use it to solve a wide range of problems in mathematics, science, and engineering.