Given the function by f(x)= ln(x^2 - 9)..?

Question

Write the formula fot the inverse function for x>3.

My question is, does x>3 make a difference?

beacause if not , my should answer be f^-1(x)=2x/(x^2 - 9),right?

Answer 1

yes, the x>3 makes a difference, but only in regards to the domain of f(x).

But no, your calculation of the inverse is not correct. Remember, to solve for the inverse, you swap x and y (or f(x) in this case) and solve for y. Here you've got a natural log. That should give you a hint that you should be using an exponential somewhere.

The function that you solved for is the inverse function of:

(sqrt(9x^2+1)+1)/x or -(sqrt(9x^2+1)-1)/x

Remember, to check if something is the inverse, substitute it back in. you should get x back.

Answer 2

Yes, the x>3 is important. It is important because the function ln(y) doesn't work if y is less than or equal to zero. Try it in your calculator; you'll get an error

In this case, if x is equal to three, you'll get ln(0), which doesn't work. Less than 3, and you'll get ln() of some negative number, which also doesn't work.

It doesn't really make a difference in your answer, though, except for the answer you only want the positive square root (not the negative square root)

And the answer should be: f^-1(x)=sqrt(e^x + 9)
(where sqrt(y) is the positive square root of y, and "e" is the number e)

You get it like this:

f(x)=ln(x^2-9)

e^f(x) = e^( ln(x^2 - 9) )

e^f(x) = x^2 - 9

e^f(x) + 9 = x^2

sqrt(e^f(x) + 9) = x

Thus, f^-1(x)=sqrt(e^x + 9)

Answer 3

The restriction that x>3 keeps ln(x^2-9) in the range of the real numbers.

Now, let y=ln(x^2-9) so e^y=x^2 - 9 and solve for x

x = √((e^y)+9)

This is why x>3 happens (for real-valued x,y) since, no matter how large y gets (in the negative direction) e^y only approaches 0 so ((e^y)+9) approaches 9 and
√((e^y)+9) approaches 3.


Doug

Answer 4

Set the function equivalent to a special variable, say "y": f(x) = y y = (x^2 - 4) / (2x^2) Now sparkling up this for x in words of y. you receives a sparkling function in words of y. This function might want to be the inverse of f(x). the way in which to target that's to likely merely multiply each and each area by technique of 2x^2 and integrate like words. you may want to finally get: (2y - a million)*x^2 = 4 x^2 = 4 / (2y-a million) x = sqrt(4 / (2y-a million)) and so on. So the inverse function of f(x), enable's call it g(y), is: g(y) = (2 * sqrt(2y-a million)) / (2y-a million)

Answer 5

Yes, it makes a difference because if x < or = to 3, your original function does not exist. You can't take the Ln of 0 or a negative number.

y = (e^x+9)^.5

That should be your inverse.

How did I get that? replace x with y and solve for your new x.

Answer 6

you need to have x^2>9, which means either x>3 or x<-3, that is,

f^-1(x) = +/- sqrt (e^x+9)

Answer 7

it makes a difference, becaue otherwise it might not be defined.
the answer is:
y=ln(x^2-9)
e^y=x^2-9
e^y+9=x^2
square root of (e^y+9)=x
so f^{-1}(x)= square root of (e^x+9)

Answer 8

Crap I used to know how to do that when I took Algebra... now I'm taking Trigonometry. Sorry I can't help!

Answer 9

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