Question

1. given f(x), find f⁻¹. f(x) = 2x² - 3

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We have the function \( f(x) = 2x^2 - 3 \). First, we rewrite it as \( y = 2x^2 - 3 \). This step is to make it easier to swap the variables \( x \) and \( y \) in the next step, which is a standard procedure for finding the inverse of a function.

Step2: Swap \( x \) and \( y \)

After swapping \( x \) and \( y \), the equation becomes \( x = 2y^2 - 3 \). Swapping the variables is based on the definition of an inverse function, where if \( y = f(x) \), then \( x = f^{-1}(y) \), and we want to express \( y \) in terms of \( x \) to get \( f^{-1}(x) \).

Step3: Solve for \( y \)

First, we isolate the term with \( y^2 \). Add 3 to both sides of the equation:
\( x + 3 = 2y^2 \)
Then, divide both sides by 2:
\( \frac{x + 3}{2} = y^2 \)
To solve for \( y \), we take the square root of both sides. Remember that when we take the square root, we have two solutions (positive and negative) because \( (\pm a)^2 = a^2 \). So we get:
\( y = \pm\sqrt{\frac{x + 3}{2}} \)

Step4: Replace \( y \) with \( f^{-1}(x) \)

Now we replace \( y \) with \( f^{-1}(x) \) to get the inverse function:
\( f^{-1}(x) = \pm\sqrt{\frac{x + 3}{2}} \)
However, we need to note the domain and range of the original function to determine if the inverse is a function. The original function \( f(x) = 2x^2 - 3 \) has a domain of all real numbers and a range of \( y \geq - 3 \). For the inverse to be a function, we need to restrict the domain of the original function. If we consider the original function with a restricted domain (e.g., \( x \geq 0 \) or \( x \leq 0 \)), the inverse will be a function. But if we consider the original function over its entire domain, it is not one - to - one (since it is a parabola opening upwards, and for \( y> - 3 \), there are two \( x \) values that map to the same \( y \) value), so the inverse is a relation (not a function) unless we restrict the domain. But following the algebraic steps to find the inverse relation, we have \( f^{-1}(x)=\pm\sqrt{\frac{x + 3}{2}} \) with the domain \( x\geq - 3 \) (since the expression inside the square root must be non - negative, \( \frac{x + 3}{2}\geq0\Rightarrow x + 3\geq0\Rightarrow x\geq - 3 \)).

Answer:

\( f^{-1}(x)=\pm\sqrt{\frac{x + 3}{2}}, \ x\geq - 3 \) (If we consider the inverse relation; if we restrict the domain of \( f(x) \) to \( x\geq0 \) or \( x\leq0 \), we can have a function - valued inverse, e.g., for \( x\geq0 \), \( f^{-1}(x)=\sqrt{\frac{x + 3}{2}}, \ x\geq - 3 \); for \( x\leq0 \), \( f^{-1}(x)=-\sqrt{\frac{x + 3}{2}}, \ x\geq - 3 \))