Question

For each of the following functions, \cdot determine the equation for the inverse, \(f^{-1}(x)\) \(\cdot \operatorname{graph} f(x)\) and \(f^{-1}(x)\).determine the domain and range of \(f(x)\) and \(f^{-1}(x)\) \(f(x)=3 x+2\) b) \(f(x)=4-2 x\) c) \(f(x)=\frac{1}{2} x-6\) d) \(f(x)=x^{2}+2, x \leq 0\) e) \(f(x)=2-x^{2}, x \geq 0\)

Short answer

Find the inverse by swapping \(x\) and \(y\) and solving for \(y\). Graph both functions and determine their domains and ranges.

Step-by-step solution

- Inverse of the function

To find the inverse of a function, swap the roles of the dependent and independent variables, and solve for the new dependent variable.

Part (a) \(f(x)=3x+2\)

1. Replace \(f(x)\) with \(y\): \(y = 3x + 2\) 2. Swap \(x\) and \(y\): \(x = 3y + 2\) 3. Solve for \(y\): \(y = \frac{x - 2}{3}\) Thus, the inverse function is: \(f^{-1}(x) = \frac{x - 2}{3}\) **Graphing**: Graph both \(y = 3x + 2\) and \(y = \frac{x - 2}{3}\) **Domain and Range**:- \(f(x)\): Domain: \( (-\infty, \infty)\), Range: \( (-\infty, \infty)\) - \(f^{-1}(x)\): Domain: \( (-\infty, \infty)\), Range: \( (-\infty, \infty)\)

Part (b) \( f(x) = 4 - 2x \)

1. Replace \( f(x) \) with \( y \): \( y = 4 - 2x \) 2. Swap \( x \) and \( y \): \( x = 4 - 2y \) 3. Solve for \( y \): \( y = \frac{4 - x}{2}\) Thus, the inverse function is: \( f^{-1}(x) = \frac{4 - x}{2} \) **Graphing**: Graph both \( y = 4 - 2x \) and \( y = \frac{4 - x}{2} \) **Domain and Range**:- \(f(x)\): Domain: \( (-\infty, \infty)\), Range: \( (-\infty, \infty)\) - \(f^{-1}(x)\): Domain: \( (-\infty, \infty)\), Range: \( (-\infty, \infty)\)

Part (c) \( f(x) = \frac{1}{2}x - 6 \)

1. Replace \( f(x) \) with \( y \): \( y = \frac{1}{2}x - 6 \) 2. Swap \( x \) and \( y \): \( x = \frac{1}{2}y - 6 \) 3. Solve for \( y \): \( y = 2x + 12\) Thus, the inverse function is: \( f^{-1}(x) = 2x + 12 \) **Graphing**: Graph both \( y = \frac{1}{2}x - 6 \) and \( y = 2x + 12 \) **Domain and Range**:- \(f(x)\): Domain: \( (-\infty, \infty)\), Range: \( (-\infty, \infty)\) - \(f^{-1}(x)\): Domain: \( (-\infty, \infty)\), Range: \( (-\infty, \infty)\)

Part (d) \( f(x) = x^2 + 2 \), where \( x \leq 0 \)

1. Replace \( f(x) \) with \( y \): \( y = x^2 + 2 \) 2. Swap \( x \) and \( y \): \( x = y^2 + 2 \) 3. Solve for \( y \): \( y = \sqrt{x - 2} \) Restricting to non-positive \( x \): \( f^{-1}(x) = -\sqrt{x - 2} \) **Graphing**: Graph both \( y = x^2 + 2 \) and \( y = -\sqrt{x - 2} \) **Domain and Range**:- \(f(x)\): Domain: \( (-\infty, 0] \), Range: \( [2, \infty)\) - \(f^{-1}(x)\): Domain: \( [2, \infty)\), Range: \((-\infty, 0]\)

Part (e) \( f(x) = 2 - x^2 \), where \( x \geq 0 \)

1. Replace \( f(x) \) with \( y \): \( y = 2 - x^2 \) 2. Swap \( x \) and \( y \): \( x = 2 - y^2 \) 3. Solve for \( y \): \( y = \sqrt{2 - x} \) Restricting to non-negative \( x \): \( f^{-1}(x) = \sqrt{2 - x} \) **Graphing**: Graph both \( y = 2 - x^2 \) and \( y = \sqrt{2 - x} \) **Domain and Range**:- \(f(x)\): Domain: \( [0, \infty)\), Range: \( (-\infty, 2] \) - \(f^{-1}(x)\): Domain: \( (-\infty, 2] \), Range: \([0, \infty)\)

Key concepts

Function Domain and Range

Understanding the domain and range of a function is critical in precalculus. The **domain** refers to all possible input values (x-values) that a function can accept. For instance, the function \(f(x) = 3x + 2\) has a domain of \((-abla, abla)\), meaning any real number can be an input. The **range** refers to all output values (y-values) a function can produce. For \(f(x) = 3x + 2\), the range is also \((-abla, abla)\). This tells us that no matter what value of \(x\) we insert into the function, it can produce any real number as \(y\). Now, let's take \(f(x) = x^2 + 2\), where \(x \le 0\). Here, the domain is restricted to \((-abla, 0]\), meaning \(x\) can only be less than or equal to zero. Correspondingly, the range is \([2, abla)\), meaning the function's output values will always be 2 or higher because squaring a negative number gives a positive result, and adding 2 shifts the range upwards.

Graphing Functions

Graphing functions gives a visual understanding of their behavior. When graphing a function such as \(f(x)=3x+2\), you plot points by inserting various values of \(x\) and finding corresponding values of \(y\). This produces a straight line with a slope of 3 and a y-intercept at 2. The inverse function \(f^{-1}(x)=\frac{x-2}{3}\), when graphed, would create a line that reflects \(f(x)\) over the line \(y = x\). Graphing both functions, you see that each point on \(f(x)\) is mirrored by its counterpart on \(f^{-1}(x)\). For non-linear functions, such as \(f(x)=2-x^2\) for \(x \ge 0\), the graph will be a downward parabola starting from (0,2). Its inverse, \(f^{-1}(x)=\sqrt{2-x}\), reflects it horizontally. When sketching these graphs, ensure the domain and range are respected to guarantee accuracy.

Precalculus

Precalculus lays the groundwork for calculus by solidifying our understanding of functions, graphs and their properties. Key concepts include the **inverse of a function**, which swaps the dependent and independent variables, and requires solving for the new dependent variable. For instance, given \(f(x) = 4 - 2x\), the inverse function \(f^{-1}(x)\) is found by first replacing \(f(x)\) with \(y\): \(y=4-2x\), swapping \(x\) and \(y\): \(x=4-2y\), and solving for \(y\): \(y=\frac{4-x}{2}\). This gives us \(f^{-1}(x)=\frac{4-x}{2}\), and it allows us to graph the inverse alongside the original function. Precalculus also involves understanding how to interpret and solve equations, inequalities, and verifies that all mappings meet the criteria for functions. Mastering these concepts ensures a smoother transition to calculus topics.