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 -\sqrtx.)

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 -\sqrtx.)

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

Answer 2

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

First, let's find the critical points of the function. These are the points where the derivative of the function is equal to zero. Taking the derivative of f(x) = 2x^2-4x-5 with respect to x, we get:

f'(x) = 4x - 4

Setting f'(x) = 0, we have:

4x - 4 = 0
4x = 4
x = 1

So, x = 1 is a critical point of the function.

Next, we can analyze the concavity of the function to determine if it has a maximum or minimum point at x=1. Taking the second derivative of f(x), we get:

f''(x) = 4

Since the second derivative is positive for all values of x, the function is concave up. This means there is a minimum point at x = 1.

Therefore, the largest interval that includes x = 0 for which the function is invertible would be (-∞, 1]. This interval starts from negative infinity and includes x = 0, but stops before reaching x = 1, the minimum point, ensuring the function is one-to-one.

The inverse function of f(x) = 2x^2-4x-5 restricted to this interval can be found by switching the x and y variables and solving for y.

Answer 3

To determine the largest interval that includes the point x=0 where the function f(x) = 2x^2-4x-5 can be made invertible, we need to find the range of the function within that interval.

Let's start by determining the range of the function f(x) = 2x^2-4x-5.

To find the range, we can consider the vertex of the parabola described by the quadratic term 2x^2-4x-5. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of the quadratic term. In this case, a = 2 and b = -4, so the x-coordinate of the vertex is given by:

x = -(-4) / (2 * 2) = 4 / 4 = 1.

So, the vertex is at the point (1, f(1)). Now, we can find the value of f(1):

f(1) = 2(1)^2 - 4(1) - 5 = 2 - 4 - 5 = -7.

This means the vertex is at the point (1, -7).

Since the coefficient in front of the x^2 term is positive (2 > 0), the parabola opens upward and the vertex represents the minimum point of the function. Therefore, the range of the function f(x) = 2x^2-4x-5 is all real numbers greater than or equal to the y-coordinate of the vertex (-7 in this case).

To make the function invertible, we need to restrict the domain to ensure that each y-value of the function corresponds to a unique x-value. In other words, we want to restrict the domain such that the function is one-to-one.

Since we want to include the point x=0, the largest interval we can choose is [0, inf). This interval includes the point x=0 and contains all real numbers greater than or equal to 0. Within this interval, the function f(x) = 2x^2-4x-5 will be invertible.

Please note that the inverse function itself is not being calculated here. This analysis simply determines the interval within which the function can be made invertible.