Some functions that aren't invertible can be made invertible by restricting their domains. For example, the function x^2 is invertible if we restrict x to the interval [0,\inf), or to any subset of that interval. In that case, the inverse function is-\sqrt x. (We could also restrict x^2 to the domain (-\inf,0], in which case the inverse function would be -\sqrt x.)

Question

Some functions that aren't invertible can be made invertible by restricting their domains. For example, the function x^2 is invertible if we restrict x to the interval [0,\inf), or to any subset of that interval. In that case, the inverse function is-\sqrt x. (We could also restrict x^2 to the domain (-\inf,0], in which case the inverse function would be -\sqrt x.)

Similarly, by restricting the domain of the function f(x) = 2x^2-4x-5 to an interval, we can make it invertible. What is the largest such interval that includes the point x=0?

Answer 1

f(x) = 2(x-1)^2 - 7

So, if we restrict the domain to x>1 or x<1 we can find an inverse

I expect you can turn that info into an interval . . .

Answer 2

To find the largest interval that includes the point x = 0, for which the function f(x) = 2x^2 - 4x - 5 is invertible, we need to determine the range of values where the function is one-to-one or has a unique inverse.

One approach is to analyze the function using calculus. We can take the derivative of f(x) to determine where its slope is positive, negative, or zero. In this case, the derivative of f(x) is given by f'(x) = 4x - 4.

Next, we look at the critical points of the function by setting f'(x) equal to zero and solving for x:
4x - 4 = 0
4x = 4
x = 1

From this calculation, we find that x = 1 is a critical point of the function.

To determine the intervals where the function is increasing or decreasing, we can choose test points within each interval and check the sign of f'(x). For example, if we choose a test point less than 1, such as x = 0, we can evaluate f'(0) = 4(0) - 4 = -4, which is negative. Similarly, if we choose a test point greater than 1, such as x = 2, we can evaluate f'(2) = 4(2) - 4 = 4, which is positive.

Based on this information, we can conclude that f(x) is decreasing for x < 1 and increasing for x > 1. Therefore, the function is one-to-one and, thus, invertible within the interval (-∞, 1) and (1, +∞).

However, we want to find the largest interval that includes the point x = 0. We know that the function is invertible for x < 1, which means that the largest interval that includes the point x = 0 is (-∞, 1].

To summarize, by restricting the domain of the function f(x) = 2x^2 - 4x - 5 to the interval (-∞, 1], we can make it invertible, and this is the largest such interval that includes the point x = 0.