Problem 51 - Algebra 2

Question

For each function \(f,\) find \(f^{-1},\) the domain and range of \(f\) and \(f^{-1},\) and determine whether \(f^{-1}\) is a function. $$ f(x)=\frac{x^{2}}{2} $$

Step-by-Step Solution

Answer

The inverse function \(f^{-1}(x)\) is \(\sqrt{2x}\). The domain of \(f(x)\) is \(-\infty, \infty\) and the range is \([0, \infty)\). The domain of \(f^{-1}(x)\) is \([0, \infty)\) and the range is \(-\infty, \infty\). Furthermore, \(f^{-1}(x)\) is indeed a function.

Step 1: Find the Inverse Function \(f^{-1}(x)\)

Start by replacing \(f(x)\) with \(y\), so the equation becomes \(y = \frac{x^{2}}{2}\). To find the inverse, we need to switch the roles of \(x\) and \(y\). Thus, the equation becomes \(x = \frac{y^{2}}{2}\). Now we'll solve for \(y\) to get the inverse function. Multiply both sides of the equation \(x = \frac{y^{2}}{2}\) by 2, that gives us \(2x = y^{2}\). Taking the square root of both sides gives us \(y = \sqrt{2x}\). So our inverse function \(f^{-1}(x) = \sqrt{2x}\)

Step 2: Determine the Domain and Range of \(f(x)\) and \(f^{-1}(x)\)

The domain of \(f(x)\) is all real numbers because we can put any real number into \(f(x) = \frac{x^{2}}{2}\) and get a real number output. Therefore, the domain of \(f(x)\) is \(-\infty, \infty\). The range of \(f(x)\) is all non-negative real numbers because the square of a real number is always non-negative. Therefore, the range of \(f(x)\) is \([0, \infty)\). Similarly, since \(f^{-1}(x)\) only involves a square root, its domain is only non-negative numbers, i.e., \([0, \infty)\) and the range of \(f^{-1}(x)\) is all real numbers as there is no restrictions on the output sign of \(f^{-1}(x)\), thus \(-\infty, \infty\).

Step 3: Determine if \(f^{-1}(x)\) is a function

A relation is a function only when each input corresponds to exactly one output. Looking at \(f^{-1}(x)\), it is clear that for each \(x\) from the domain \([0, \infty)\), there is exactly one output. Hence, \(f^{-1}(x)\) is a function.

Key Concepts

Domain and Range

Understanding the domain and range of a function is crucial in mathematics. The domain refers to all the possible input values for the function, which are essentially the values you can plug into the function to get a valid output. For the function \( f(x) = \frac{x^2}{2} \), the domain is all real numbers, \( \mathbb{R} \), because squaring any real number and dividing by two yields another real number.
The range, on the other hand, refers to all the possible output values the function can produce. Here, since squaring a number always results in a non-negative value, the range of \( f(x) \) is \([0, \infty)\).

• The domain of \( f(x) \) is \((-\infty, \infty)\).
• The range of \( f(x) \) is \([0, \infty)\).
• For the inverse function \( f^{-1}(x) = \sqrt{2x} \), the domain is restricted to \([0, \infty)\).
• The range of \( f^{-1}(x) \) is \((-\infty, \infty)\) as the square root function can produce any real number output from its domain.

Quadratic Functions

Quadratic functions are a type of polynomial that play a key role in algebra. They are characterized by the presence of an \( x^2 \) term and are usually in the form \( ax^2 + bx + c \). These functions have a distinctive parabolic graph shape.
The function provided here, \( f(x) = \frac{x^2}{2} \), is a simple quadratic function with no linear or constant term, making it a pure parabola through the origin. It opens upwards since the coefficient of \( x^2 \) is positive.
Quadratic functions can always be graphed to determine their vertex and direction of opening. In this case:

• The vertex of this function is at the origin (0,0).
• Since the coefficient is positive, the parabola opens upwards.
• The axis of symmetry is the y-axis, \( x = 0 \).

These properties help define the behavior and limitations of the function, particularly when considering its inverse.

Function Validity

Determining if a relation is a valid function requires ensuring that each input has exactly one output. This is the essence of the vertical line test: if a vertical line crosses a graph more than once at any point, the relation is not a function.
In the case of the inverse of our function \( f(x) = \frac{x^2}{2} \), the function \( f^{-1}(x) = \sqrt{2x} \) passes this test. Each non-negative x-value from the domain \([0, \infty)\) corresponds to exactly one output, making \( f^{-1} \) a valid function.
However, note that not every quadratic function will have a valid inverse function. Only functions that are one-to-one over their entire domain can have inverses that remain functions.

• Ensure the original function is one-to-one by restricting its domain if necessary.
• Verify the inverse function using function tests like the horizontal line test on the original before finding the inverse.