Angelica is working with function machines. She has the two machines shown at right. She wants to put them in order so that the output of the first machine becomes the input of the second. Her initial input is
a. In what order must she put the machines to get a final output of 5 ? order:
b. Is it possible for her to find an initial input that will get a final output of
? If so, show how she could do that. If not, explain why not.
The Correct Answer and Explanation is:
Let’s solve the problem step-by-step using the information provided in the image.
The two function machines shown are:
- y=x−5y = \sqrt{x – 5}y=x−5
- y=x2−6y = x^2 – 6y=x2−6
Part a: In what order must she put the machines to get a final output of 5?
We are given an initial input of 6 and need a final output of 5.
Let’s test both possible orders:
Order 1: First use y=x−5y = \sqrt{x – 5}y=x−5, then y=x2−6y = x^2 – 6y=x2−6
- Start with input 6:
- First machine: 6−5=1=1\sqrt{6 – 5} = \sqrt{1} = 16−5=1=1
- Second machine: 12−6=1−6=−51^2 – 6 = 1 – 6 = -512−6=1−6=−5 → not 5
Order 2: First use y=x2−6y = x^2 – 6y=x2−6, then y=x−5y = \sqrt{x – 5}y=x−5
- Start with input 6:
- First machine: 62−6=36−6=306^2 – 6 = 36 – 6 = 3062−6=36−6=30
- Second machine: 30−5=25=5\sqrt{30 – 5} = \sqrt{25} = 530−5=25=5 → this works
✅ Correct Order:
- First use y=x2−6y = x^2 – 6y=x2−6
- Then use y=x−5y = \sqrt{x – 5}y=x−5
Part b: Is it possible for her to find an initial input that will get a final output of -5?
Let’s examine whether it’s possible to get –5 as the final output.
Case 1: Final machine is y=x−5y = \sqrt{x – 5}y=x−5
This function always gives non-negative outputs because a square root of a real number cannot be negative.
So if the second machine is y=x−5y = \sqrt{x – 5}y=x−5, the final output can never be –5.
Case 2: Final machine is y=x2−6y = x^2 – 6y=x2−6
Let’s see if we can make this equal to –5: x2−6=−5×2=1⇒x=±1x^2 – 6 = -5 \\ x^2 = 1 \Rightarrow x = \pm 1×2−6=−5×2=1⇒x=±1
So, the input to this machine must be 1 or –1. That means the output of the first machine must be 1 or –1.
Try solving:
- For the first machine y=x−5y = \sqrt{x – 5}y=x−5, can it output 1 or –1?
We check:
- x−5=1⇒x−5=1⇒x=6\sqrt{x – 5} = 1 \Rightarrow x – 5 = 1 \Rightarrow x = 6x−5=1⇒x−5=1⇒x=6 → OK
- x−5=−1\sqrt{x – 5} = -1x−5=−1 → invalid because square roots can’t be negative
So only x=6x = 6x=6 works.
✅ Therefore, if the machine order is:
- First: y=x−5y = \sqrt{x – 5}y=x−5
- Then: y=x2−6y = x^2 – 6y=x2−6
- Input: 6 → Output: 1 → Final Output: 12−6=−51^2 – 6 = -512−6=−5
✅ Final Answers:
a. Order:
First use y=x2−6y = x^2 – 6y=x2−6, then y=x−5y = \sqrt{x – 5}y=x−5
b. Yes, it’s possible to get a final output of –5.
Use the order:
First y=x−5y = \sqrt{x – 5}y=x−5, then y=x2−6y = x^2 – 6y=x2−6,
and start with input x=6x = 6x=6.
