Respuesta :

For [tex]f(-20)[/tex], you start by determining which of the pieces the x-value of -20 fits into.

Wow, those are some convoluted conditions on the right.  

Note: I originally had this post with using x = 20 and not x = -20.  The rest of this has been corrected to reflect that change.

Eventually you see "x<-20" on the third line/third piece of the function.  So that tells you that you need to use the third piece and for x=-20, just treat the function as is it was

    [tex]f(x) = 3x-7[/tex]

This means,

    [tex]f(-20) = 3(-20)-7[/tex]

Can you finish that out from there?

Recap:

Step #1: figure out which piece your x-value fits with.

Step #2: plug in the x-value into that piece and evaluate it.

Answer: See below

Step-by-step explanation:

For this problem, we are given four functions, and their domains. All we have to do is figure out which equation we want to plug x into and solve.

f(-20)

For this first problem, we have x=-20. Then we can eliminate equations 1, 2, 4. We know equation 3 is the correct one because it says x<-12. -20 is less than -12. Therefore we plug in x=-20 to the third equation.

f(-20)=3(-20)-7              [multiply]

f(-20)=-60-7                  [subtract]

f(-20)=-67

--------------------------------------------------------------------------------------------

f(0)

For this problem, we have x=0. Then we can eliminate equations 1, 2, 3. We know equation 4 is the correct one because it says x∈{-15, -12, 0, 10}. This means x has to be either one of the four numbers listed in the parentheses. We see that x=0 is in there. Therefore we plug in x=0 to the fourth equation.

f(0)=11-|-0-1|             [subtract]

f(0)=11-|-1|                [absolute value]

f(0)=11-1                    [subtract]

f(0)=10

--------------------------------------------------------------------------------------------

f(15)

For this problem, we have x=15. Then we can eliminate equations 2, 3, 4. We know equation 1 is the correct one because it says x>10. This means x has to be greater than 10. 15 is greater than 10. Therefore we plug in x=15 to the first equation.

[tex]f(15)=\frac{1}{|15+1|-11}[/tex]       [add]

[tex]f(15)=\frac{1}{|16|-11}[/tex]           [absolute value]

[tex]f(15)=\frac{1}{16-11}[/tex]             [subtract]

[tex]f(15)=\frac{1}{5}[/tex]