## whadduptori Group Title 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 one year ago one year ago

1. whpalmer4

Aren't you supposed to find the inverse?

im just so confuse.... i have no idea what im doing

3. whpalmer4

Is it written like this? $f(x) = (3-4x)^2, find f^{-1}(x)$

yes!

5. 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?

ohh1! nope i honestly do not remember

wait you switch the x and y right?

8. 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 = <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

10. whpalmer4

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))

12. whpalmer4

that's half of it...

13. whpalmer4

Won't 1/4 (3 + sqrt(x)) = y also?

yes so it would be y = (1/4)( 3 ± sqrt(x))

15. whpalmer4

Right. So, is the inverse a function?

f-1(x) = y = (1/4)(3 ± sqrt(x)) ??

oh no sorry its not a function!

18. 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.

19. 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.

20. whpalmer4

21. whpalmer4

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!! (:

23. whpalmer4

Glad I could help make the little bulb come to life :-)

24. whpalmer4

little light bulb that is

hahah you are awesome! can you maybe help me with one more?

26. whpalmer4

I can try!

Use the Change of Base Formula to evaluate log7(76)

like would it be log7(76)= log(76)/log10(7) is that the answer?

29. whpalmer4

$\log_b a = \frac{\log a }{\log b}$ So $\log_7 76 = \frac{\log{76}}{\log 7}$

30. 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?

wait so would it be log76 = 1.88081359 and log 7= 84509804 and you would divide those right?

32. whpalmer4

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

35. 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 $\log u^n = n \log u$

36. 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.

so my answer was right?

38. whpalmer4

Yes!

ah thank you again!!!!!

40. 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 :-)

I think I will! you are amazing thank you again, and I usually get confused but you made me understand!! :)

$\sqrt{7}(\sqrt{x}-7\sqrt{7}$

43. whpalmer4

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.

45. whpalmer4

Well, if you distribute it out, what do you get?

$\sqrt{7x}-49\sqrt{x}$

47. whpalmer4

where did the \sqrt{x} next to the 49 come from?

i have no idea.. how do you distribute that?

49. 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?

50. whpalmer4

and that last bit reduces to -49

51. 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).

52. whpalmer4

$a(b-c) = ab - ac, a = \sqrt{7}, b=\sqrt{x}, c = 7\sqrt{7}$

wait whats the answer.. im lost

54. 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