Question

If \(f: R \rightarrow R\) defined by \(f(x)=x^{4}+2\) then the value of \(f^{-1}(83)\) and \(\mathrm{f}^{-1}(-2)\) respectively are. (a) \(\Phi,\\{3,-3\\}\) (b) \(\\{3,-3\\}, \Phi\) (c) \(\\{4,-4\\}, \Phi\) (d) \(\\{4,-4\\},\\{2,-2\\}\)

Short answer

The short answer is (b) \(\{3, -3\},\ \Phi\).

Step-by-step solution

Find the inverse function

To find the inverse function \(f^{-1}(x)\) of the given function \(f(x) = x^4 + 2\), we first need to follow these steps: 1. Replace \(f(x)\) with \(y\): \(y = x^4 + 2\) 2. Swap \(x\) and \(y\): \(x = y^4 + 2\) 3. Solve for \(y\): \(y = \pm\sqrt[4]{x - 2}\) Now we have found the inverse function, which is \(f^{-1}(x) = \pm\sqrt[4]{x - 2}\).

Find \(f^{-1}(83)\)

To find the value of \(f^{-1}(83)\), we need to substitute \(x = 83\) into the inverse function and solve for \(y\): \(f^{-1}(83) = \pm\sqrt[4]{83 - 2} = \pm\sqrt[4]{81}\) Since both \(3^4 = 81\) and \((-3)^4 = 81\), the value of \(f^{-1}(83)\) is \(\{3, -3\}\).

Find \(f^{-1}(-2)\)

To find the value of \(f^{-1}(-2)\), we need to substitute \(x = -2\) into the inverse function and solve for \(y\): \(f^{-1}(-2) = \pm\sqrt[4]{-2 - 2} = \pm\sqrt[4]{-4}\) Since there is no real number that could make the expression inside the fourth root positive, the value of \(f^{-1}(-2)\) does not exist. We represent this with the symbol \(\Phi\). Therefore, the correct answer is (b) \(\{3, -3\},\ \Phi\).

Key concepts

Function Composition

Function composition involves combining two or more functions to form a new function. It is like putting functions together, one after another. This means applying one function to the results of another function.

For functions \( g(x) \) and \( f(x) \), the composite function is written as \((f \circ g)(x)\), which means \( f(g(x)) \). You first apply \( g(x) \), then apply \( f \) to the result.

In our example, we dealt with functions and their inverses. An inverse function basically "undoes" the effect of a function. If \( f(x) \) changes \( a \) to \( b \), then \( f^{-1}(x) \) changes \( b \) back to \( a \).

• Let \( f(x) = x^4 + 2 \).
• Its inverse is \( f^{-1}(x) = \pm\sqrt[4]{x-2} \).

By using inverse functions, we are reversing the effect of the original function, like our task of determining \( f^{-1}(83) \) and \( f^{-1}(-2) \). The first reversed to \( \{3, -3\} \) because applying \( f(x) \) to these values returns 83. For \( f^{-1}(-2) \), the expression isn't valid in real numbers, indicated by \( \Phi \).

Real Numbers

Real numbers include every number that can be found on the number line. This includes both rational and irrational numbers.

Real numbers can be positive, negative, or zero. They include:

• Integers like -3, 0, 14
• Fractions like \( \frac{1}{2}, \frac{3}{4} \)
• Irrational numbers like \( \sqrt{2}, \pi \)

In our exercise, we used real numbers to evaluate the inverse function. For example, when calculating \( f^{-1}(83) \), we solved \( \pm\sqrt[4]{81} \), which resulted in real numbers \( 3 \) and \( -3 \).

When we tried to solve \( f^{-1}(-2) \), no real number could satisfy the condition, meaning it has no real solution. Therefore, \( \Phi \) is used to express the absence of a real number answer.This shows how real numbers play a crucial role in defining the possibility of what our inverse functions can output.

Complex Numbers

Complex numbers extend real numbers by including imaginary numbers. They are written in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. \( i \) is the imaginary unit with the property \( i^2 = -1 \).

When real numbers fall short, for instance, when taking a root of a negative number, complex numbers step in to offer solutions. If you try to find a real number solution for \( \pm\sqrt[4]{-4} \) from our example, you turn to complex numbers.

• For \( \pm\sqrt[4]{-4} \), it indicates values like \( \pm\sqrt[4]{4i^2} \), producing complex results.

Understanding complex numbers can extend your problem-solving toolkit. With complex numbers, even when real number solutions don't exist, solutions can often be found in the complex plane.

This is especially useful for advanced mathematics and engineering, where complex numbers provide insights beyond the capacity of regular real numbers. Yet, in our context of evaluating inverse functions, as there are no valid real results for \( f^{-1}(-2) \), we stick with \( \Phi \) as the interpretive solution.