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!

TealSeal May 20, 2021

#1**-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

Guest May 20, 2021

#3**+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.

=^._.^=

catmg May 20, 2021