Question

Calculus homework! Find the inverse of...

Answer 1

\[e ^{\sqrt{x}/4}\]

Answer 2

\[x \ge0\]

Answer 3

Please observe FIRST, the Domain of \(f(x) = e^{\sqrt{x}/4}\). It is hoped you decide on \(x \ge 0\).

Swap and solve. You haven't done this, before calculus?

Answer 4

i did but i forgot it all.

Answer 5

@tkhunny so I sap x for 0?

Answer 6

No, the Domain is to remind you what you are doing. With \(x \ge 0\), you had better end up with a RANGE of \(y \ge 0\) for the inverse function.

You have \(y = e^{\sqrt{x}/4}\)

Swap: \(x = e^{\sqrt{y}/4}\)

Solve for y.

Answer 7

oh i seee

Answer 8

BTW - This really is not encouraging. This is NOT calculus material. You should have this well-fixed in your mind before you see calculus. This is just be official warning that you could struggle quite a bit if your algebra is not up to speed. You may wish to conduct a more formal review, just to make sure you survive.

Answer 9

@tkhunny did not come for the lecture, dude. this first hw is meant to be a review. it's still a calc class.

Answer 10

There's always a lecture. My intent is to help you succeed. I'm just telling you the truth. If you need some additional review, you should do it formally. Hoping to pick it up as you go along is very unlikely to contribute to a successful calculus experience. I've seen this pattern over and over and over. You have to get up on your algebra, trigonometry, and geometry in order to have a particularly successful calculus experience. You can succeed without a particularly good background in these things, but it is a much greater struggle.

Answer 11

go away!

Answer 12

What did you get for the inverse?

Answer 13

\[(4\ln x)^{2}\]

Answer 14

@tkhunny so is the domain \[x >0 ?\]

Answer 15

\(y = e^{\sqrt{x}/4}\)

Domain: \(x \ge 0\)
Range: \(y \ge 1\)

It is important to know where these come from. It takes a little thought on the Range.

IF our task is to create an inverse FUNCTION, this information is of utmost importance. When we find such a function, it MUST relate tot eh original function in this very specific way:

Range: \(x \ge 0\)
Domain: \(y \ge 1\)

I just switched the words.

Your algebra was fine. We have a candidate for the inverse FUNCTION.

\(y = (4\cdot ln(x))^{2}\)

The really big question is this: Does it match the given criteria for an Inverse Function? In other words, what are the Domain and Range of the Relation you have suggested is the inverse? Let's see if it qualifies.

Answer 16

Whoops, I forgot to switch x and y in the new definition.

Should be:

Domain: \(x \ge 1\) <== This is the tricky one!
Range: \(y \ge 0\)

Sorry about that.

Answer 17

@tkhunny I'm gonna say this info. thank you.

Answer 18

I'm guessing that anyone else in you class will overlook Domain issues. They will submit:

\(f^{-1}(x) = (4\cdot ln(x))^{2}\) as the Inverse Function.

The correct answer for an Inverse Function is:

\(f^{-1}(x) = (4\cdot ln(x))^{2}\;for\;x \ge 1\)

There is a funny piece on \(0 \le x \lt 1\) that just doesn't belong!