Problem 40
Question
Find the inverse function of \(f,(\mathbf{b})\) graph \(f\) and \(f^{-1}\) on the same set of coordinate axes, ( \(\mathbf{c} )\) describe the relationship between the graphs, and ( \(\mathbf{d} )\) state the domain and range of \(f\) and \(f^{-1} .\) \(f(x)=x^{2}, \quad x \geq 0\)
Step-by-Step Solution
Verified Answer
The inverse function of \(f(x) = x^2\) is \(f^{-1}(x) = \sqrt{x}\). The graph of the original function is a right half of an upward parabola and the graph of the inverse function is a right half of a sideways parabola on the first quadrant. They are mirror images reflected over \(y = x.\) Both functions have the same domain and range, \(x \geq 0\) and \(y \geq 0\) respectively.
1Step 1: Finding the Inverse Function
In finding the inverse of a function, the usual approach is to switch \(x\) and \(y\) and solve for \(y\). \(x = y^2\) Now, solve it for \(y\) to find the inverse function: \(f^{-1}(x) = \sqrt{x}\).
2Step 2: Graphing the Functions
The function \(f(x) = x^2\) is a parabola opening upwards. In this case, since \(x \geq 0\), the graph will only include the right half of the parabola. The inverse function \(f^{-1}(x) = \sqrt{x}\) is half of a sideways-opening parabola occupying the first quadrant.
3Step 3: Describing the Relationship
The graph of \(f^{-1}(x)\) is a reflection of the graph of \(f(x)\) in the line \(y = x\). Both graphs intersect this line at the same points.
4Step 4: Stating the Domain and Range
The domain of \(f(x) = x^2\) is \(x \geq 0\) and its range is \(y \geq 0\). The domain of its inverse function \(f^{-1}(x)\) is \(x \geq 0\) and its range is \(y \geq 0\) as well.
Key Concepts
Understanding ParabolasExploring Domain and RangeReflection in Line \(y = x\)
Understanding Parabolas
Parabolas have a unique and familiar shape that resembles a U or an upside-down U, also known as head "opening." In our case, the function \( f(x) = x^2 \) represents a parabola that opens upwards. When graphing this function, you'll notice it is symmetric, meaning both sides look identical on a vertical line passing through its vertex.
- The vertex is the point \((0,0)\) in this scenario, which serves as the minimum point of the parabola.
- This parabola is special because it's restricted to \(x \geq 0\), focusing only on the right half.
Exploring Domain and Range
The domain of a function concerns the possible input values, while the range handles the possible outputs. For \( f(x) = x^2 \), where \( x \geq 0 \), the function's domain consists of non-negative numbers. This means it includes 0 and all positive numbers.
For the range, we look at potential outputs, which, because the graph shows only the right half of the parabola, include all values \(y \geq 0\). Essentially, parabola output cannot be negative.
For the range, we look at potential outputs, which, because the graph shows only the right half of the parabola, include all values \(y \geq 0\). Essentially, parabola output cannot be negative.
- The domain of \(f\) is \(x \geq 0\).
- The range of \(f\) is \(y \geq 0\).
- Interestingly, these same values flip for the inverse \(f^{-1}(x) = \sqrt{x}\).
Reflection in Line \(y = x\)
Visualizing the relationship between a function and its inverse is fascinating and critical for comprehension. The process creates a reflection over the line \(y = x\). This line acts as a mirror, ensuring each point on the original function finds a corresponding mirrored point on the inverse function.
For example, a point \((a,b)\) on \(f\) will show as \((b,a)\) on \(f^{-1}\). In our function \(f(x) = x^2\), consider the point \((4,16)\); its inverse point is \((16,4)\), reflecting over \(y=x\).
For example, a point \((a,b)\) on \(f\) will show as \((b,a)\) on \(f^{-1}\). In our function \(f(x) = x^2\), consider the point \((4,16)\); its inverse point is \((16,4)\), reflecting over \(y=x\).
- Such a reflection creates a symmetrical pattern where each graph intersects at points where \(f(x) = x\) and \(f^{-1}(x) = x\).
- This understanding departs from the simple task of finding inverse equations and cleverly demonstrates the concept graphically.
Other exercises in this chapter
Problem 40
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