Problem 27
Question
Find an explicit formula for \(f^{-1}\) and use it to graph \(f^{-1}, f,\) and the line \(y=x\) on the same screen. To check your work, see whether the graphs of \(f\) and \(f^{-1}\) are reflections about the line. $$ f(x)=x^{4}+1, \quad x \geqslant 0 $$
Step-by-Step Solution
Verified Answer
The inverse function is \(f^{-1}(x) = \sqrt[4]{x - 1}\). The graphs are reflections across the line \(y = x\).
1Step 1: Understand the Function
The function given is \( f(x) = x^4 + 1 \) with the domain \( x \geq 0 \). This is a one-to-one function due to the limitation on its domain.
2Step 2: Swap Variables
To find the inverse of \( f(x) \), start by swapping \( x \) and \( y \) in the equation \( y = x^4 + 1 \). This gives us \( x = y^4 + 1 \).
3Step 3: Solve for y in terms of x
Isolate \( y \) in the equation \( x = y^4 + 1 \) by first subtracting 1 from both sides, resulting in \( x - 1 = y^4 \). Then take the fourth root of both sides to solve for \( y \). Thus, \( y = \sqrt[4]{x - 1} \).
4Step 4: Define the Inverse Function
The inverse function is then \( f^{-1}(x) = \sqrt[4]{x - 1} \) with the domain \( x \geq 1 \), which corresponds to the range of the original function \( f(x) \).
5Step 5: Graph the Functions
Graph \( f(x) = x^4 + 1 \), \( f^{-1}(x) = \sqrt[4]{x - 1} \), and the line \( y = x \). Ensure that \( f(x) \) and \( f^{-1}(x) \) are reflections about the line \( y = x \). Confirm that the graphs intersect the line \( y = x \) at the points \((1,1)\) due to symmetry.
Key Concepts
One-to-One FunctionGraphing Inverse FunctionsReflection about the Line y=x
One-to-One Function
In mathematics, a function is called "one-to-one" or injective if each output value is paired with precisely one input value. This means that no two distinct inputs produce the same output. It is like having a sturdy lock-and-key system where each key can open only one specific lock.
For our function \(f(x) = x^4 + 1\), it is defined over the domain \(x \geq 0\). This makes it behave as a one-to-one function within this range. When graphed, each point on the curve corresponds uniquely to a single input value. The significance of a function being one-to-one is crucial, especially when finding an inverse. Without this property, the function could not be inverted perfectly, and some values would be mapped ambiguously.
For our function \(f(x) = x^4 + 1\), it is defined over the domain \(x \geq 0\). This makes it behave as a one-to-one function within this range. When graphed, each point on the curve corresponds uniquely to a single input value. The significance of a function being one-to-one is crucial, especially when finding an inverse. Without this property, the function could not be inverted perfectly, and some values would be mapped ambiguously.
Graphing Inverse Functions
Graphing inverse functions means plotting both the original function and its inverse on the same coordinate plane. This allows for a visual understanding of how these functions relate to each other.
When we have a one-to-one function like \( f(x) = x^4 + 1 \), its inverse can indeed be found by swapping the variables \(x\) and \(y\) and solving for \(y\). Thus, for \( f^{-1}(x) \), the formula derived is \( f^{-1}(x) = \sqrt[4]{x - 1} \).
When graphing these, ensure the domain and range match appropriately. The domain of the inverse function is \( x \geq 1 \), corresponding to the range of the original function. Plotting these will show how they interact as two halves of a relationship: where one injects inputs to outcomes, and the other retraces outcomes back to inputs.
When we have a one-to-one function like \( f(x) = x^4 + 1 \), its inverse can indeed be found by swapping the variables \(x\) and \(y\) and solving for \(y\). Thus, for \( f^{-1}(x) \), the formula derived is \( f^{-1}(x) = \sqrt[4]{x - 1} \).
When graphing these, ensure the domain and range match appropriately. The domain of the inverse function is \( x \geq 1 \), corresponding to the range of the original function. Plotting these will show how they interact as two halves of a relationship: where one injects inputs to outcomes, and the other retraces outcomes back to inputs.
Reflection about the Line y=x
One of the most fascinating properties of inverse functions is their geometric reflection across the line \(y = x\). When plotting \(f(x) = x^4 + 1\) and its inverse \(f^{-1}(x) = \sqrt[4]{x - 1}\), they should mirror each other across this diagonal line.
Imagine the line \(y = x\) as a perfect mirror. Every point \((a, b)\) on the original function should correspond to a point \((b, a)\) on the inverse function when reflected. This symmetrical relationship visually confirms that one function is truly the inverse of the other.
When drawn accurately, their graphs should intersect the line \(y = x\) at exactly one point, for instance, at \((1, 1)\) in this case. This symmetry not only helps validate your calculations but also enhances the understanding of inverse functions and their inherent characteristics.
Imagine the line \(y = x\) as a perfect mirror. Every point \((a, b)\) on the original function should correspond to a point \((b, a)\) on the inverse function when reflected. This symmetrical relationship visually confirms that one function is truly the inverse of the other.
When drawn accurately, their graphs should intersect the line \(y = x\) at exactly one point, for instance, at \((1, 1)\) in this case. This symmetry not only helps validate your calculations but also enhances the understanding of inverse functions and their inherent characteristics.
Other exercises in this chapter
Problem 27
\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply,
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Find the derivative of the function. Simplify where possible. \(y=x \sin ^{-1} x+\sqrt{1-x^{2}}\)
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Find the limit. $$\lim _{x \rightarrow 2^{+}} e^{3 /(2-x)}$$
View solution Problem 28
Differentiate the function. $$ y=\frac{e^{u}-e^{-u}}{e^{u}+e^{-u}} $$
View solution