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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

 Dec 31, 2020
 #1
avatar+96 
+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. :)))

=^._.^=

 Dec 31, 2020
 #2
avatar+312 
0

Thank you so much. I really appreciate your help. I understand the problem a lot better now.

 Dec 31, 2020

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