Let f be defined by

\(f(x) = \left\{ \begin{array}{cl} 2-x & \text{ if } x \leq 1, \\ 2x-x^2 & \text{ if } x>1. \end{array} \right.\)

Calculate \(f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)\).

Thank you

imdumb Dec 31, 2020

#1**+1 **

Hello imdumb! (I'm sure you're actually a humble genius)

Inverses are f(a) = b, f^(-1)(b) = a. I'm sorry, I can't figure out how to get the -1 thingy on latex.

**Equation 1: **

Let's start with f^(-1)(-3) = x.

So f(x) = -3.

We'll go through both cases.

**Case 1: **

If f(x) = 2 - x when x =< 1

Then x must be 5 (2 - 5 = -3), however this isn't possible because x needs to less than or equal to 1.

**Case 2: **

If f(x) = 2x - x^2 when x > 1

Then x = -1 or 3 (solving the quadratic 2x - x^2 = -3), however, x > 1 so x = 3.

**Answer: **

Thus, f^(-1)(-3) = 3.

**Equation 2: **

Moving on to the next one, f^(-1)(0) = x.

So f(x) = 0.

**Case 1: **

If f(x) = 2 - x when x =< 1

Then x must be 2, but not possible since 2 is not less than or equal to 1.

**Case 2:**

If f(x) = 2x - x^2 when x > 1

Then x = 2 or x = -2, but x > 1 so x = 2.

**Answer: **

Thus, f^(-1)(0) = 2.

**Equation 3: **

Finally we have, f^(-1)(3) = x

So f(x) = 3.

**Case 1: **

If f(x) = 2 - x when x =< 1

Then x must be -1, which works perfectly since 1 =< 1.

If case 1 is correct, we don't have to test case 2.

**Answer: **

Thus, f^(-1)(3) = -1

**Final Answer:**

Now we just need to add everything.

3 + 2 - 1 = 4

Hopefully 4 is the correct answer.

I hope this helped. :)))

=^._.^=

catmg Dec 31, 2020