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+1
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avatar+106 

Let\[f(x) = \left\{\begin{array}{cl}-x + 3 & \text{if } x \le 0, \\2x - 5 & \text{if } x > 0.\end{array}\right.\]How many solutions does the equation $f(f(x)) = 4$ have?

 

Thanks!angel

 May 20, 2021
 #1
avatar
-2

I think if you go on youtube and search "Never Gonna Give You up" and click the first video, it should have the answer

 May 20, 2021
 #2
avatar+106 
+1

cheekyhaha lol, but I really need helpwink

TealSeal  May 20, 2021
 #3
avatar+1823 
+1

f(x) = 4

The first case, -x + 3, which would be x = -1. 

This works since -1 =< 0. 

The second case, 2x - 5, x = 4.5. 

This works since 4.5 > 0. 

 

f(x) = -1

The first case -x + 3, x = 2. 

However this does not work since 2 is not less than or equal to 0. 

The second case 2x - 5, x = 2. 

This works since 2 > 0. 

 

f(f(2)) = 4

 

f(x) = 4.5

The first case -x + 3, x = -1.5.

This works since -1.5 =< 0

The second case 2x - 5, x = 9.5/2. 

Thiw works since 9.5/2 > 0. 

 

f(f(-1.5)) = 4

f(f(4.75)) = 4

 

So we have 3 cases.

 

=^._.^=

 May 20, 2021
 #5
avatar+106 
+1

TYSM! It's correcct! laugh

TealSeal  May 21, 2021
 #6
avatar+1823 
+1

You're welcome. :))

 

=^._.^=

catmg  May 21, 2021
 #4
avatar
+2

I got 3          2, -3/2, and 19/4

Do you know the right answer?

 May 21, 2021
 #7
avatar+1823 
+1

Nice work. :)

 

=^._.^=

catmg  May 21, 2021

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