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

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- whpalmer4

Aren't you supposed to find the inverse?

- anonymous

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

- whpalmer4

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

yes!

- 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

ohh1! nope i honestly do not remember

- anonymous

wait you switch the x and y right?

- 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

ok so in my example would it be..
y = (3 – 4x)²
x = (3-4y)^2

- whpalmer4

so after the dust settles, what does y = ?

- anonymous

so x = (3-4y)^2
y= 1/4 (3-sqrt(x))
so... f^-1 = 1/4 (3-sqrt(x))

- whpalmer4

that's half of it...

- whpalmer4

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

- anonymous

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

- whpalmer4

Right. So, is the inverse a function?

- anonymous

f-1(x) = y = (1/4)(3 ± sqrt(x)) ??

- anonymous

oh no sorry its not a function!

- 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

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

##### 1 Attachment

- 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

oh okay! i see what your saying now! you explained it to me well! thank you for your help!! (:

- whpalmer4

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

- whpalmer4

little light bulb that is

- anonymous

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

- whpalmer4

I can try!

- anonymous

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

- anonymous

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

- whpalmer4

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

- 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

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

- whpalmer4

Yes, but I think you are missing a decimal in your value of log 7...

- anonymous

oh yeah i did! so it would .84509804

- anonymous

so the answer would be 2.22556685

- 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

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

so my answer was right?

- whpalmer4

Yes!

- anonymous

ah thank you again!!!!!

- 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

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

- anonymous

\[\sqrt{7}(\sqrt{x}-7\sqrt{7}\]

- whpalmer4

Is that supposed to be \[\sqrt{7}(\sqrt{x}-7\sqrt{7})\] and is the goal to simplify it?

- anonymous

oh yes sorry! Multiply and simplify if possible.

- whpalmer4

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

- anonymous

\[\sqrt{7x}-49\sqrt{x}\]

- whpalmer4

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

- anonymous

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

- 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

and that last bit reduces to -49

- 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

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

- anonymous

wait whats the answer.. im lost

- 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

Oh I see what you were doing now! sorry

- whpalmer4

Not a problem. If we were talking, I would have said a bit more, and you would have understood right away.

Looking for something else?

Not the answer you are looking for? Search for more explanations.