Hello!
Actually, this function g has no (one-valued) inverse function. For some y's there is no such x that g(x)=y, for some there are two such x's (two solutions of a quadratic equation). But we can consider two-valued functions, too.
The function ax^2+bx+c with a positive a has its minimum at x_0 = (-b)/(2a). For our function g it is -(-4)/4 = 1, the minimum value is g(1) = 2 - 4 - 5 = -7. So there are two points x_1 and x_2 such that g(x) = -2. To find them, we have to solve the quadratic equation
2x^2-4x-5=-2, or 2x^2-4x-3=0.
The solutions are x_(1,2)=(2+-sqrt(2^2+2*3))/2 = (2+-sqrt(10))/2 = 1+-sqrt(5/2).
The final answer depends on the additional conditions. One may state that there are no g^(-1)(-2), that there are two values, 1-sqrt(5/2) and 1+sqrt(5/2), or choose one of them if there are some constraints on the domain of g.
Saturday, February 17, 2018
Given g(x)= 2x^2-4x-5, find the following, g^(-1)(-2).
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