Juliana has created the function f(x)=3x+2/4 to represent the cost of texting on her current plan, where x represents the number of texts. Juliana discovers that using the inverse function to solve for x=24, she can predict how many texts she can use for $24. Explain to Juliana how to accomplish this, using complete sentences
@Gerardo_cast23 @Falco276 @Hero
Is it \[f(x) = \frac{3x + 2}{4}\] or \[f(x) = 3x + \frac{2}{4}\]
the first one!
What's interesting is, you actually have to find the inverse of the function.
I suppose that's what you're having trouble with, right?
YES exactly
To find the inverse of the function, you're supposed to change \(f(x)\) to \(y\), swap \(x\) and \(y\), then isolate y afterwards.
So you have \[f(x) = \frac{3x + 2}{4}\] Changing \(f(x)\) to \(y\) you get: \[y = \frac{3x + 2}{4}\] Swapping \(x\) and \(y\) you get: \[x = \frac{3y + 2}{4}\] Can you isolate \(y\) from here?
thats where I have the problemm
To finish, 1. Multiply 4 on both sides 2. Subtract 2 on both sides 3. Divide both sides by 3
i dont get where u multiply the 4 on the fraction side
\[4 \dot\ x = \frac{3y + 2}{4} \dot\ 4\]\[4 \dot\ x = \frac{3y + 2}{\cancel{4}} \dot\ \cancel{4}\]\[4x = 3y + 2\]
y = 4x-2 / 3
Lastly you have to replace \(y\) with \(f^{-1}(x)\)
f -1 (x) = y= 4x-2 /3
Now, that you have inversed the function, the domain and range are the inverse of what they were before.
Thank you! Can u help with another pleasE?
oh but now my answer..do i sub 24?
I was just about to ask you about that
You have to find \(f^{-1}(24)\)
So 4(24) -2 / 3?
@Hero
Yes, what did you get?
31.3
Well, we should say approximately 31 texts for $24
Since it is not possible to send .3 texts.
lol yeah ok so 31 texts she can send
Correct.
To check, input f(31) into the original function to see if you get 24.
Ok . please help me understand this one
Marrie is logging the number of radioactive isotope she observes every minute. She has determined her function to be f(x) =-4x +120. Based on the situatuion above describe and neccesary restrictions to the domain and range
I've answered this kind of question already. You should scroll through my questions answered to find it.
okkay !
u have way too many lol
Thank you @broskishelleh
@Hero You're a saint! Thanks for clearing that up!
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