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Linear functions are simple functions expressed in the form of \( f(x) = ax + b \), where \( a \) and \( b \) are constants. These functions graph as straight lines, and they have a unique property: every linear function has an inverse, as long as the slope \( a \) is not zero.
To find the inverse of a linear function, you swap \( x \) and \( y \) (where \( y = f(x) \)) and solve for the new \( x \). For example, given \( f(x) = 2x - 3 \):

• Let \( y = 2x - 3 \)
• Solve for \( x \): \( y + 3 = 2x \) yielding \( x = \frac{y + 3}{2} \)
• Thus, the inverse function is \( f^{-1}(x) = \frac{x + 3}{2} \)

The inverse function, \( f^{-1}(x) \), effectively undoes the original function. So if you apply \( f \) followed by \( f^{-1} \) to any input within the domain, you'll get your original input back!
Linear functions, due to their one-to-one nature, don't require domain restrictions to have inverses, making them straightforward in this context.

Rational functions are expressions of the form \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials, and \( q(x) eq 0 \). Finding inverses for rational functions can be a bit more complex due to division by variables and potential domain restrictions.
Consider \( f(x) = \frac{2x - 3}{x + 4} \):

• Set \( y = \frac{2x - 3}{x + 4} \)
• Cross-multiply to simplify: \( y(x + 4) = 2x - 3 \)
• Rearrange to isolate \( x \): \( x = \frac{-3 - 4y}{y - 2} \)
• So, \( f^{-1}(x) = \frac{-3 - 4x}{x - 2} \)

In rational functions, always look out for values that make the denominator zero. Here, the function \( f(x) \) excludes \( x = -4 \), and the inverse \( f^{-1}(x) \) excludes \( x = 2 \). These exclusions ensure both the function and inverse are defined across their respective domains.
Rational functions have inverses when they are one-to-one over their domain, signifying the output is unique to each input.

Quadratic functions are functions of the form \( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants. These functions graph as parabolas, which are symmetric and open either upwards or downwards. However, they are not one-to-one across their natural domain, making them unfit for a straightforward inverse.
When you try to find an inverse, you notice that both positive and negative roots overlap, as shown in \( f(x) = x^2 + 1 \). For an inverse function, it's crucial to have a unique mapping from output to input.
To correct this, restrict the domain. If we limit our domain to \( x \geq 0 \), the function becomes one-to-one, effectively removing the overlap.

• From \( y = x^2 + 1 \), solve for \( x \): \( x = \sqrt{y - 1} \)
• The inverse is \( f^{-1}(x) = \sqrt{x - 1} \) but only valid for \( x \geq 1 \). This corresponds to the range of \( f(x) \) on the restricted domain.

By restricting the domain of a quadratic function, it allows us to define an inverse function that truly reflects its original structure over a well-behaved section.