How do you find the inverse of the function f(x) = 3x^2 + 4 when x = 2?

To find the inverse of the function f(x) = 3x² + 4, we start by replacing f(x) with y:

y = 3x² + 4

Next, we solve this equation for x in terms of y:

1. Subtract 4 from both sides:

y – 4 = 3x²

2. Divide by 3:

(y – 4)/3 = x²

3. Take the square root of both sides:

x = ±√((y – 4)/3)

The function is not one-to-one for all real numbers because it involves a squared term. To ensure we have a valid inverse, we restrict the domain of f(x). Let’s assume we are considering the function for x ≥ 0. Thus, we will only take the positive root:

x = √((y – 4)/3)

Now, to express the inverse function, we switch x and y:

f^-1(x) = √((x – 4)/3)

Finally, we can evaluate this inverse function at x = 2:

f^-1(2) = √((2 – 4)/3) which simplifies to f^-1(2) = √(-2/3). Since the result is imaginary, it indicates that x = 2 is outside the range of the function given the restriction.

In conclusion, you cannot find a real value for the inverse function at x = 2 with the function f(x) = 3x² + 4 when adhering to the domain of non-negative reals.