Question

Let \( f(x) = 2x^3 - 3x \neq 4 \). Then the inverse function \( f^{-1}(x) \) is:

Hint:

To find the inverse of a function, swap \( x \) and \( y \) and solve for \( y \). Ensure the function is one-to-one for the inverse to exist.

Answer 1

The given function is \( f(x) = 2x^3 - 3x \neq 4 \). The inequality sign seems to be a typo; it’s likely meant to be \( f(x) = 2x^3 - 3x + 4 \). We proceed with this interpretation. 
To find the inverse \( f^{-1}(x) \), set \( y = f(x) \): \[ y = 2x^3 - 3x + 4 \] 
Solve for \( x \) in terms of \( y \): \[ 2x^3 - 3x + 4 - y = 0 \implies 2x^3 - 3x + (4 - y) = 0 \] 
This is a cubic equation in \( x \), which is complex to solve analytically. Let’s test if the function is invertible by checking if it’s one-to-one (monotonic). Compute the derivative: \[ f'(x) = 6x^2 - 3 = 3(2x^2 - 1) \] 
Roots of \( f'(x) = 0 \): \( 2x^2 - 1 = 0 \implies x = \pm \frac{1}{\sqrt{2}} \). Since \( f'(x) \) changes sign, \( f(x) \) is not monotonic everywhere, so it’s not globally invertible. 
However, the options suggest a linear inverse, indicating the function might be different. Let’s assume a simpler function due to the linear options. Suppose the intended function was linear, but \( 2x^3 - 3x + 4 \) is clearly cubic. 
Reinterpret: If the function was meant to be linear (e.g., a typo), let’s try a linear function that fits the options. Test option (D) by assuming \( f(x) = \frac{2x + 3}{x + 3} \) (derived from the inverse form): 
If \( f^{-1}(x) = \frac{x - 3}{2x - 3} \), set \( y = \frac{x - 3}{2x - 3} \), solve for \( x \): \[ y = \frac{x - 3}{2x - 3} 
⇒ y(2x - 3) = x - 3 
⇒ 2xy - 3y = x - 3 
⇒ 2xy - x = 3y - 3 
⇒ x(2y - 1) 
= 3(y - 1)
⇒ \(x = \frac{3(y - 1)}{2y - 1}\)