Question

Consider the function g(x) = 1 / (1 + eˣ). (a) Find g⁻¹(x). (b) Determine the domain and range of g. (c) Find the value of x such that g(x) = g⁻¹(x)?

Answer 1

(a) The inverse function
`\(g^(-1)(x)\) for \(g(x) = (1)/(1 + e^x)\)` is
`\(g^(-1)(x) = \ln\left((1)/(x) - 1\right)\)`. (b) The domain of g is
`\((- \infty, \infty)\)`, and the range is 0, 1. (c) The value of x such that
`\(g(x) = g^(-1)(x)\)` is x = -1.

Step-by-step explanation:

(a) To find the inverse function
`\(g^(-1)(x)\)`, we interchange x and y in the original function and solve for y For
`g(x) = (1)/(1 + e^x)\)`, after algebraic manipulation, we get \
`(g^(-1)(x) = \ln\left((1)/(x) - 1\right)\)`.

(b) The domain of g is
`\((- \infty, \infty)\)` because the exponential function
`\(e^x\)` is defined for all real x, and the denominator
`\(1 + e^x\)` is always nonzero. The range of g is 0, 1 because
`\(g(x)\)`approaches 0 as x approaches
`\(\infty\)`and approaches 1 as x approaches
`\(-\infty\).`

(c) To find the value of x such that
`\(g(x) = g^(-1)(x)\)`, we set
`\(g(x) = \ln\left((1)/(x) - 1\right)\)` and solve for x. The solution is x = -1.