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Chapter 7: Problem 20

Gives a formula for a function \(y=f(x)\) and shows the graphs of \(f\) and \(f^{-1} .\) Find a formula for \(f^{-1}\) in each case. $$f(x)=x^{2}, \quad x \leq 0$$

Short Answer

Expert verified
The inverse function is \(f^{-1}(x) = -\sqrt{x}\).

Step by step solution

01

Understand the Given Function

We are given the function \(f(x) = x^2\) with the domain restriction \(x \leq 0\). This means we consider only the left half of the parabola, where \(x\) is non-positive.
02

Set Up the Equation for the Inverse

To find the inverse function \(f^{-1}(x)\), swap the roles of \(x\) and \(y\) in the function \(y = f(x)\). This gives us the equation \(x = y^2\) since \(y = x^2\) originally.
03

Solve for y in Terms of x

From the equation \(x = y^2\), we solve for \(y\). Given the domain restriction \(x \leq 0\), \(y\) will also be non-positive. Therefore, we take the negative square root: \(y = -\sqrt{x}\).
04

Write the Inverse Function

Thus, the inverse function is \(f^{-1}(x) = -\sqrt{x}\). This inverse function satisfies the condition \(x \leq 0\) for the original function.
05

Verify the Inverse Function

Let's verify by plugging \(f^{-1}(f(x))\) and \(f(f^{-1}(x))\) to ensure we get \(x\) back.For \(f(f^{-1}(x)) = ((-\sqrt{x})^2) = x\), it holds since \(x \leq 0\).For \(f^{-1}(f(x)) = -\sqrt{x^2} = x\), it holds as we take the negative root for \(x \leq 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabolas
Parabolas are fundamental shapes in mathematics, often represented by quadratic functions like the one presented in the exercise: \(f(x) = x^2\). A parabola is a U-shaped curve with distinct properties:
  • The vertex is its highest or lowest point, depending on its orientation.
  • It opens upwards if the coefficient of \(x^2\) is positive, and downwards if it's negative.
  • The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves.
In this exercise, we focus on the left half of the parabola because of the domain restriction \(x \leq 0\). This implies considering only the portion where the x-values are non-positive. The vertex here would logically be at the origin, given the starting point of \(x = 0\). By understanding these properties, one can learn how quadratic functions behave and how modifications鈥攍ike domain restrictions鈥攁lter their graphs.
Domain and Range
Domain and range are two critical concepts in understanding functions and their inverses. The domain of a function is all the possible values that the input \(x\) can take, while the range is what the function outputs or covers on the y-axis.
  • The given function \(f(x) = x^2\) has a domain of \(x \leq 0\), meaning we only consider non-positive values of \(x\).
  • Its corresponding range is \([0, \, \infty)\), covering all non-negative y-values, as squaring any number results in a non-negative number.
  • For the inverse \(f^{-1}(x) = -\sqrt{x}\), the domain is restricted to \(x \leq 0\) because inverses swap the domain and range of the original function.
  • The range of \(f^{-1}(x)\) becomes \((-\infty, \, 0]\) since we take only the non-positive square roots.
Clearly establishing these sets for domain and range ensures each function is properly defined and accurately graphed.
Square Root Function
Square root functions involve operations that make functions invertible by reversing the squaring action of a number. In the context of the given problem, the inverse function \(f^{-1}(x) = -\sqrt{x}\) incorporates the concept of square roots.
  • The square root operation allows you to retrieve original numbers that were squared, thus finding the inverse.
  • In our problem, we apply a negative sign because the original function \(f(x) = x^2\) is defined in a domain where \(x \leq 0\), meaning any number plugged in is non-positive.
  • The function \(y = -\sqrt{x}\) is graphically represented as a half-parabola on the left quadrant, accurately reflecting these transformations.
Recognizing how square roots are used in inverse functions is crucial for solving problems related to them.

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