Do your best i need it quickly sorta, if you do a weird answer just for the points i will report you.



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
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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]