piecewise function composition chaos!

[HighSchoolDx]

I have a piecewise function, which goes as follows:

f(x) =

x^2 - 4, FOR x >= -4
x + 3, OTHERWISE

I am trying to find, for how many values of x is it true that f(f(x)) = 5.

Currently, I have tried composing the function onto itself, and I am at

(x^2 - 4)^2 - 4 = x^4 - 8x^2 + 12.
(x + 3) + 3 = x + 6.

so then, I set these equal to 5. for the first equation, i have 4 solutions: +- 1 and +- sqrt7.
for the second, there is one solution only, x = 1.

are there 4 + 1 = 5 values of x for f(f(x)) = 5? i am not sure if i'm composing the functions correctly, or arriving at an incorrect solution. i am not very confident in piecewise functions... could you please provide a verification? if this is wrong, please tell me what next steps I should take.

Thank you for assistance!

 

[Deleted member 4993]

I have a piecewise function, which goes as follows:

f(x) =

x^2 - 4, FOR x >= -4
x + 3, OTHERWISE

I am trying to find, for how many values of x is it true that f(f(x)) = 5.

Currently, I have tried composing the function onto itself, and I am at

(x^2 - 4)^2 - 4 = x^4 - 8x^2 + 12.
(x + 3) + 3 = x + 6.

so then, I set these equal to 5. for the first equation, i have 4 solutions: +- 1 and +- sqrt7.
for the second, there is one solution only, x = 1.

are there 4 + 1 = 5 values of x for f(f(x)) = 5? i am not sure if i'm composing the functions correctly, or arriving at an incorrect solution. i am not very confident in piecewise functions... could you please provide a verification? if this is wrong, please tell me what next steps I should take.

Thank you for assistance!

You wrote [for f(f(x)):

(x + 3) + 3 = x + 6

Then you should have:

x + 6 = 5 → x = -1

I did not check your work for the other function. Please correct the above and recheck your answer.

 

[Dr.Peterson]

I have a piecewise function, which goes as follows:

f(x) =

x^2 - 4, FOR x >= -4
x + 3, OTHERWISE

I am trying to find, for how many values of x is it true that f(f(x)) = 5.

Currently, I have tried composing the function onto itself, and I am at

(x^2 - 4)^2 - 4 = x^4 - 8x^2 + 12.
(x + 3) + 3 = x + 6.

so then, I set these equal to 5. for the first equation, i have 4 solutions: +- 1 and +- sqrt7.
for the second, there is one solution only, x = 1.

are there 4 + 1 = 5 values of x for f(f(x)) = 5? i am not sure if i'm composing the functions correctly, or arriving at an incorrect solution. i am not very confident in piecewise functions... could you please provide a verification? if this is wrong, please tell me what next steps I should take.

Thank you for assistance!

Don't forget that if x<-4, f(x) might be >=-4 so the second application of the function would use the other case.

I would write the composite function as three cases (or more, if I got that wrong) resulting from that thinking, and only then look for when f(f(x)) = 5, case by case. Keep in mind that the solutions you get for x have to be in the right pieces of the composite function.

 

[JeffM]

[MATH]x \ge - 4 \implies f(x) = x^2 - 4 \implies f(f(x)) = f(x^2 - 4) = (x^2 - 4)^2 - 4 = x^4 - 8x^2 + 12.[/MATH]
Why do you question this? You decided that f(f(x)) = 5 is true when x is not less than -4 if and only if x is plus or minus the square root of 7 or plus or minus 1. Why not check your work.?

[MATH]x= \pm \sqrt{7} \implies (\pm \sqrt{7})^4 - 8(\pm \sqrt{7})^2 + 12 = 49 - 56 + 12 = 5. \ \checkmark[/MATH]
[MATH]x = \pm 1 \implies (\pm 1)^4 - 8(\pm 1)^2 + 12 = 1 - 8 + 12 = 5. \ \checkmark[/MATH]
A quartic can have at most four zeroes.

[MATH]x < - 4 \implies f(x) = x + 3 \implies f(f(x)) = f(x + 3) = (x + 3) + 3 = x + 6.[/MATH]
Why do you question this? You decided that f(f(x)) = 5 when x is less than - 4 if and only if x = 1. Why not check?

[MATH]1 + 6 = 7 \ne 5. \ \times.[/MATH]
That was wrong. If it had been right how do you count

[MATH]-1, \ - \sqrt{7}, \ \sqrt{7}, \text { and } 1[/MATH] as five solutions?

But of course that was wrong. if you solve the equation x + 6 = 5, then x = -1. But that is impossible because this equation is true only if x < -4.

Check your work, do not double count, and remember your constraints.

 

[Steven G]

I have a piecewise function, which goes as follows:

f(x) =

x^2 - 4, FOR x >= -4
x + 3, OTHERWISE

I am trying to find, for how many values of x is it true that f(f(x)) = 5.

You are computing f of something (that something just happens to be f(x)) and you are getting 5.
First task is to find out what you need to take f of to get 5?
2 + 3 = 5, but you do not use the x+ 3 rule if x>=-4 and 2 is >=-4.

So we consider the other rule now. When does x^2 - 4 = 5? That happens when x^2=9 or x= +/- 3. Note that both 3 and -3 are >=-4.

So we now know that we can only take f of 3 or -3 to get 5. Recall that we are taking f of f(x). So we need to find the x-value or x-values, if any, that make f(x) = 3 or -3.

Now it is your turn to finish up.

 

[Cubist]

@HighSchoolDx please post back with your work. There's one solution that hasn't (yet) been mentioned in this thread.

EDIT: I like the approach proposed by @Jomo

 

[lex]

There is no need to compose functions

[MATH]x+3[/MATH] is negative when [MATH]x<-4[/MATH]so [MATH]f(x)=5 \leftrightarrow x^2-4=5[/MATH][MATH]\leftrightarrow x=\pm 3\hspace1ex[/MATH] (both of which are ≥-4)

So [MATH]f(f(x))=5 \leftrightarrow f(x)=\pm 3[/MATH][MATH]\leftrightarrow[/MATH](a) [MATH]x^2-4=\pm 3 \hspace1ex \text{and }x≥-4[/MATH] or
(b) [MATH]x+3=\pm 3 \hspace1ex \text{and }x<-4[/MATH]
(a) gives the 4 solutions you have found (all of which are: [MATH]x≥-4[/MATH])
(b) gives the 5th solution (which is [MATH]<-4[/MATH])

 

[HighSchoolDx]

THANK YOU SO MUCH GUYS!!
Thanks for the helping thoughts. I figured it out from them!

Following Jomo's approach: letting y = f(x), so then f(y) = 5. --> x^2 - 4 = 5 or x + 3 = 5.

first equation x^2 - 4 = 5 --> y = +3, or -3. both >= -4
second equation x + 3 = 5 --> y = 2 --> discard this as it is not < -4.

so f(x) = +- 3 --> x^2 - 4 = +-3 or x + 3 = +-3.

first equation x^2 - 4 = +-3 --> x = +-1, x = +-sqrt7. all valid >= -4
second equation x + 3 = +-3 --> x = -6, 0. only -6 < -4.
solutions: +1, -1, +sqrt7, -sqrt7, -6
5 solutions!

Thank you once again