Problem 18
Question
Use algebra to find the inverse of the given one-to-one function. $$f(x)=1 / \sqrt{x}$$
Step-by-Step Solution
Verified Answer
Answer: The inverse of the given one-to-one function is $$f^{-1}(x) = \left(\frac{1}{x}\right)^2$$.
1Step 1: Replace f(x) with y
Write the function using y instead of f(x):
$$y = \frac{1}{\sqrt{x}}$$
2Step 2: Swap x and y
Interchange x and y in the equation:
$$x = \frac{1}{\sqrt{y}}$$
3Step 3: Solve for y
To solve for y, let's first get rid of the denominator by cross-multiplying:
$$x\sqrt{y} = 1$$
Now, divide both sides by x:
$$\sqrt{y} = \frac{1}{x}$$
Finally, square both sides to isolate y:
$$y = \left(\frac{1}{x}\right)^2$$
4Step 4: Replace y with the inverse function notation
Replace y with $$f^{-1}(x)$$ to denote the inverse function:
$$f^{-1}(x) = \left(\frac{1}{x}\right)^2$$
Thus, the inverse of the given one-to-one function $$f(x) = \frac{1}{\sqrt{x}}$$ is $$f^{-1}(x) = \left(\frac{1}{x}\right)^2$$.
Key Concepts
One-to-One FunctionsSolving EquationsFunction NotationAlgebraic Manipulation
One-to-One Functions
A one-to-one function is a specific type of function where each output value (or y-value) pairs with exactly one input value (or x-value) and vice versa. In simpler terms, no two different inputs can produce the same output. This characteristic ensures that each y-value is unique to a corresponding x-value, making it possible to reverse the function.
- Ensures unique input-output relationship
- Necessary for finding inverses
Solving Equations
Solving equations is a fundamental concept in algebra. It involves finding the value of the unknown variable that makes an equation true. When determining the inverse of a function, you often need to swap variables and solve for the new dependent variable.
- Substitute known values to verify solutions
- Techniques include isolating variables and rearranging terms
Function Notation
Function notation is a way to express operations in mathematics, often using symbols such as $f(x)$. This format highlights the relationship between inputs (x) and outputs (f(x)). Instead of using generic y, function notation offers a clearer way to denote specific functions and their inverse counterparts.
- Simplifies communication of relationships
- Denotes inverse functions with $f^{-1}(x)$
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying algebraic expressions to solve equations or find inverses. This often includes operations like adding, subtracting, multiplying, dividing, and even squaring terms to isolate variables.
- Rearrange equations to solve for a specific variable
- Essential for isolating terms to find inverses
Other exercises in this chapter
Problem 17
Find the average rate of change of the function f over the given interval. $$f(x)=\frac{x^{2}-3}{2 x-4} \text { from } x=3 \text { to } x=8$$
View solution Problem 17
Find \((g \circ f)(3),(f \circ g)(1),\) and \((f \circ f)(0)\) $$f(x)=3 x-2, \quad g(x)=x^{2}$$
View solution Problem 18
Compute and simplify the difference quotient of the function. $$f(x)=x+5$$
View solution Problem 18
Find a single viewing window that shows complete graphs of the functions \(f, g,\) and \(h.\) $$\begin{aligned}&f(x)=\left|x^{2}-5\right| ; \quad g(x)=f(x+8)\\\
View solution