Problem 20
Question
If \(f: R \rightarrow R\) is defined by \(f(x)=x^{3}\) then \(f^{-1}(8)\) is equal to \mathrm{\\{} [ K a r n a t a k a ~ C E T - 2 0 0 8 ] ~ (a) \(\\{2,-2\\}\) (b) \(\\{2,2\\}\) (c) \(\\{2\\}\) (d) \(\left\\{2,2 \omega, 2 \omega^{2}\right\\}\)
Step-by-Step Solution
Verified Answer
The answer is (c) \{2\}.
1Step 1: Understand the Function and Its Inverse
The function given is \( f(x) = x^3 \), which is a one-to-one function. Therefore, it has an inverse. The inverse function \( f^{-1}(x) \) will give the value of \( x \) when \( f(x) = y \). Our goal is to find \( f^{-1}(8) \).
2Step 2: Set Up the Equation for Inversion
To find \( f^{-1}(8) \), we need to set \( f(x) = 8 \). This means solving the equation \( x^3 = 8 \).
3Step 3: Solve the Equation
The equation \( x^3 = 8 \) can be solved by taking the cube root of both sides. This yields \( x = \sqrt[3]{8} \).
4Step 4: Calculate the Cube Root
The cube root of 8 is calculated as \( x = 2 \) because \( 2^3 = 8 \).
5Step 5: Interpret the Result
The function \( f(x) = x^3 \) is strictly increasing on all real numbers, indicating it is a one-to-one function. Therefore, the only solution is \( x = 2 \).
Key Concepts
One-to-one functionsCube RootsReal Numbers
One-to-one functions
Understanding one-to-one functions is crucial to grasping inverse functions. A one-to-one function, also known as an injective function, is a function where every element of the range is mapped to exactly one element of the domain. This means that if you have two different inputs, they will produce two different outputs. Because of this property, each output value directly corresponds to only one input value.
To test if a function is one-to-one, you can use the Horizontal Line Test. If any horizontal line crosses the graph of the function at most once, then the function is one-to-one.
For the function \( f(x) = x^3 \), you can see that it is one-to-one because it passes the Horizontal Line Test. This property is significant because only one-to-one functions have inverses that are also functions. Thus, when you invert a one-to-one function, like \( f(x) = x^3 \), its inverse \( f^{-1}(x) \) is also a valid function.
To test if a function is one-to-one, you can use the Horizontal Line Test. If any horizontal line crosses the graph of the function at most once, then the function is one-to-one.
For the function \( f(x) = x^3 \), you can see that it is one-to-one because it passes the Horizontal Line Test. This property is significant because only one-to-one functions have inverses that are also functions. Thus, when you invert a one-to-one function, like \( f(x) = x^3 \), its inverse \( f^{-1}(x) \) is also a valid function.
Cube Roots
Cube roots are the values that, when multiplied by themselves three times, give you the original number. The cube root of a number \( n \) is denoted as \( \sqrt[3]{n} \). For example, the cube root of 8 is \( 2 \) because \( 2^3 = 8 \).
Unlike square roots, cube roots are defined for all real numbers, both positive and negative. This is because multiplying two negative numbers together yields a positive number, but multiplying three negative numbers together results in a negative number, allowing negative results under the cube root.
To solve the problem \( x^3 = 8 \), you take the cube root of both sides. The cube root process reverses the cubing, giving you \( x = \sqrt[3]{8} = 2 \). This simplicity in calculation is why cube roots are so important in solving cubic equations.
Unlike square roots, cube roots are defined for all real numbers, both positive and negative. This is because multiplying two negative numbers together yields a positive number, but multiplying three negative numbers together results in a negative number, allowing negative results under the cube root.
To solve the problem \( x^3 = 8 \), you take the cube root of both sides. The cube root process reverses the cubing, giving you \( x = \sqrt[3]{8} = 2 \). This simplicity in calculation is why cube roots are so important in solving cubic equations.
Real Numbers
Real numbers are a fundamental concept in mathematics, encompassing all the numbers that can be found on the number line. This includes integers, fractions, and irrational numbers like \( \pi \) and \( \sqrt{2} \). Real numbers are denoted by the symbol \( \mathbb{R} \).
One of the vital characteristics of real numbers is that they are ordered. This means you can compare any two real numbers and determine which is greater or lesser, or if they are equal. Understanding real numbers is essential for calculus, algebra, and many other areas of mathematics.
In the context of this exercise, understanding that the function \( f(x) = x^3 \) is defined over all real numbers is key. It means that \( f \) can take any real number as input and produce a real number as output. This all-encompassing nature of real numbers ensures that solutions to equations involving powers and roots, like \( x^3 = 8 \), are consistent and reliable across the entirety of the number line.
One of the vital characteristics of real numbers is that they are ordered. This means you can compare any two real numbers and determine which is greater or lesser, or if they are equal. Understanding real numbers is essential for calculus, algebra, and many other areas of mathematics.
In the context of this exercise, understanding that the function \( f(x) = x^3 \) is defined over all real numbers is key. It means that \( f \) can take any real number as input and produce a real number as output. This all-encompassing nature of real numbers ensures that solutions to equations involving powers and roots, like \( x^3 = 8 \), are consistent and reliable across the entirety of the number line.
Other exercises in this chapter
Problem 18
Which one of the following function is oneto-one? \(\quad\) [Kerala PET-2008] (a) \(f(x)=\sin x, x \in[-\pi, \pi)\) (b) \(f(x)=\sin x, x \in\left[\frac{-3 \pi}{
View solution Problem 19
If \(f(x)=2 x+1\) and \(g(x)=\frac{x-1}{2}\) for all real \(x\) then \((\operatorname{fog})^{-1}\left(\frac{1}{x}\right)\) is equal to \mathrm{\\{} [ K e r a l
View solution Problem 17
If \(f(x)=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}}\), then \(f^{-1}(x)\) is (a) \(\frac{1}{2} \log _{2} \frac{x}{1-x}\) (b) \(\frac{1}{2} \log _{2} \frac{1+x}{1-x}\) c
View solution