anonymous
  • anonymous
Can somebody just check me? For the function f(x) = (3 – 4x)², find f–1. Determine whether f–1 is a function. Is this right? what i have: f(x)=(3-4x)^2 f(1)=(3-4x1)^2 =(3-4)^2 =(-1)^2 =1 f is a function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
whpalmer4
  • whpalmer4
Aren't you supposed to find the inverse?
anonymous
  • anonymous
im just so confuse.... i have no idea what im doing
whpalmer4
  • whpalmer4
Is it written like this? \[f(x) = (3-4x)^2, find f^{-1}(x)\]

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anonymous
  • anonymous
yes!
whpalmer4
  • whpalmer4
Okay, yeah, that's finding the inverse function and checking if that is a function, not evaluating the original function at f(1) or f(-1)... So, do you know how to find the inverse function?
anonymous
  • anonymous
ohh1! nope i honestly do not remember
anonymous
  • anonymous
wait you switch the x and y right?
whpalmer4
  • whpalmer4
Okay, if you want to find the inverse function, you write out the original function as y = f(x) = (3-4x)^2, but then you swap x and y, and rearrange it as y = For example, if you had y = f(x) = 3x, f^(-1)(x) would be found like this: y = 3x x = 3y x/3 = y so y = x/3 would be the inverse of y = 3x.
anonymous
  • anonymous
ok so in my example would it be.. y = (3 – 4x)² x = (3-4y)^2
whpalmer4
  • whpalmer4
so after the dust settles, what does y = ?
anonymous
  • anonymous
so x = (3-4y)^2 y= 1/4 (3-sqrt(x)) so... f^-1 = 1/4 (3-sqrt(x))
whpalmer4
  • whpalmer4
that's half of it...
whpalmer4
  • whpalmer4
Won't 1/4 (3 + sqrt(x)) = y also?
anonymous
  • anonymous
yes so it would be y = (1/4)( 3 ± sqrt(x))
whpalmer4
  • whpalmer4
Right. So, is the inverse a function?
anonymous
  • anonymous
f-1(x) = y = (1/4)(3 ± sqrt(x)) ??
anonymous
  • anonymous
oh no sorry its not a function!
whpalmer4
  • whpalmer4
That's right, it fails the vertical line test. A vertical line passes through the curve more than once, so there is not a unique value of y for every x.
whpalmer4
  • whpalmer4
If you draw a line through the origin heading up and to the right at a 45 degree angle (slope = 1), the inverse function is simply the original function reflected along that line. So if you have a parabola opening upward centered on the y-axis, the inverse is a parabola opening to the right centered on the x-axis.
whpalmer4
  • whpalmer4
whpalmer4
  • whpalmer4
And a little thought about that suggests that functions more complicated than straight lines may not have an inverse which is a function...
anonymous
  • anonymous
oh okay! i see what your saying now! you explained it to me well! thank you for your help!! (:
whpalmer4
  • whpalmer4
Glad I could help make the little bulb come to life :-)
whpalmer4
  • whpalmer4
little light bulb that is
anonymous
  • anonymous
hahah you are awesome! can you maybe help me with one more?
whpalmer4
  • whpalmer4
I can try!
anonymous
  • anonymous
Use the Change of Base Formula to evaluate log7(76)
anonymous
  • anonymous
like would it be log7(76)= log(76)/log10(7) is that the answer?
whpalmer4
  • whpalmer4
\[\log_b a = \frac{\log a }{\log b}\] So \[\log_7 76 = \frac{\log{76}}{\log 7}\]
whpalmer4
  • whpalmer4
Both of the logs on the right hand side need to be in the same base, of course. Do you know how you can show that this is true?
anonymous
  • anonymous
wait so would it be log76 = 1.88081359 and log 7= 84509804 and you would divide those right?
whpalmer4
  • whpalmer4
Yes, but I think you are missing a decimal in your value of log 7...
anonymous
  • anonymous
oh yeah i did! so it would .84509804
anonymous
  • anonymous
so the answer would be 2.22556685
whpalmer4
  • whpalmer4
\[\log_b a = \frac{\log a}{\log b}\]Raise both sides as powers of b \[b^{\log_b a} = b^{\frac{\log a}{\log b}}\]But the left half of that is simply a by definition \[a = b^{\frac{\log a}{\log b}}\]and if we take the log of both sides \[\log a = \frac{\log a}{\log b}*\log b\]because \[\log u^n = n \log u\]
whpalmer4
  • whpalmer4
As a quick check of your answer, 7^2*(7*1/5) = 68.6, and 7^2*(7*1/4) = 85 so it looks like the answer is in the right ballpark. 7^2.225 is a little bit bigger than 1/5, and a little smaller than 1/4.
anonymous
  • anonymous
so my answer was right?
whpalmer4
  • whpalmer4
Yes!
anonymous
  • anonymous
ah thank you again!!!!!
whpalmer4
  • whpalmer4
Now you can go tutor all of your friends :-) That's a good way to reinforce your grasp of the material — try to explain it to someone else. Works best if they can ask questions in return, so explaining it to the cat or dog might not be as effective :-)
anonymous
  • anonymous
I think I will! you are amazing thank you again, and I usually get confused but you made me understand!! :)
anonymous
  • anonymous
\[\sqrt{7}(\sqrt{x}-7\sqrt{7}\]
whpalmer4
  • whpalmer4
Is that supposed to be \[\sqrt{7}(\sqrt{x}-7\sqrt{7})\] and is the goal to simplify it?
anonymous
  • anonymous
oh yes sorry! Multiply and simplify if possible.
whpalmer4
  • whpalmer4
Well, if you distribute it out, what do you get?
anonymous
  • anonymous
\[\sqrt{7x}-49\sqrt{x}\]
whpalmer4
  • whpalmer4
where did the \sqrt{x} next to the 49 come from?
anonymous
  • anonymous
i have no idea.. how do you distribute that?
whpalmer4
  • whpalmer4
Well, multiply sqrt(7) by the first term in the parentheses, and you get sqrt(7)sqrt(x). Multiply sqrt(7) by the second term in the parentheses, and you get -sqrt(7)*7*sqrt(7), right?
whpalmer4
  • whpalmer4
and that last bit reduces to -49
whpalmer4
  • whpalmer4
\[\sqrt{7x}-49 = \sqrt{7}\sqrt{x}-49\] (I'd probably write the first one if it was a final answer, the second if I was going to do further canceling or manipulation).
whpalmer4
  • whpalmer4
\[a(b-c) = ab - ac, a = \sqrt{7}, b=\sqrt{x}, c = 7\sqrt{7}\]
anonymous
  • anonymous
wait whats the answer.. im lost
whpalmer4
  • whpalmer4
See my post with the (I'd probably write ...) The last bit was just showing why the second term shouldn't have an x in it
anonymous
  • anonymous
Oh I see what you were doing now! sorry
whpalmer4
  • whpalmer4
Not a problem. If we were talking, I would have said a bit more, and you would have understood right away.

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