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whadduptori
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
Aren't you supposed to find the inverse?
im just so confuse.... i have no idea what im doing
Is it written like this? \[f(x) = (3-4x)^2, find f^{-1}(x)\]
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?
ohh1! nope i honestly do not remember
wait you switch the x and y right?
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 = <some function> 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.
ok so in my example would it be.. y = (3 – 4x)² x = (3-4y)^2
so after the dust settles, what does y = ?
so x = (3-4y)^2 y= 1/4 (3-sqrt(x)) so... f^-1 = 1/4 (3-sqrt(x))
Won't 1/4 (3 + sqrt(x)) = y also?
yes so it would be y = (1/4)( 3 ± sqrt(x))
Right. So, is the inverse a function?
f-1(x) = y = (1/4)(3 ± sqrt(x)) ??
oh no sorry its not a function!
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.
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.
And a little thought about that suggests that functions more complicated than straight lines may not have an inverse which is a function...
oh okay! i see what your saying now! you explained it to me well! thank you for your help!! (:
Glad I could help make the little bulb come to life :-)
little light bulb that is
hahah you are awesome! can you maybe help me with one more?
Use the Change of Base Formula to evaluate log7(76)
like would it be log7(76)= log(76)/log10(7) is that the answer?
\[\log_b a = \frac{\log a }{\log b}\] So \[\log_7 76 = \frac{\log{76}}{\log 7}\]
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?
wait so would it be log76 = 1.88081359 and log 7= 84509804 and you would divide those right?
Yes, but I think you are missing a decimal in your value of log 7...
oh yeah i did! so it would .84509804
so the answer would be 2.22556685
\[\log_b a = \frac{\log a}{\log b}\]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 \[\log u^n = n \log u\]
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.
so my answer was right?
ah thank you again!!!!!
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 :-)
I think I will! you are amazing thank you again, and I usually get confused but you made me understand!! :)
\[\sqrt{7}(\sqrt{x}-7\sqrt{7}\]
Is that supposed to be \[\sqrt{7}(\sqrt{x}-7\sqrt{7})\] and is the goal to simplify it?
oh yes sorry! Multiply and simplify if possible.
Well, if you distribute it out, what do you get?
\[\sqrt{7x}-49\sqrt{x}\]
where did the \sqrt{x} next to the 49 come from?
i have no idea.. how do you distribute that?
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?
and that last bit reduces to -49
\[\sqrt{7x}-49 = \sqrt{7}\sqrt{x}-49\] (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).
\[a(b-c) = ab - ac, a = \sqrt{7}, b=\sqrt{x}, c = 7\sqrt{7}\]
wait whats the answer.. im lost
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
Oh I see what you were doing now! sorry
Not a problem. If we were talking, I would have said a bit more, and you would have understood right away.