# Homework Help: Inverse shminverse

1. Feb 18, 2010

### srfriggen

1. The problem statement, all variables and given/known data

If f(x)= x^3 +4x + 6,

a.) show f(x) is one to one.

b) Find inverse f(10)... f^-1(10) (hard to write type on a computer)

c) Find f^-1(10)'

2. Relevant equations

f^-1(x)' = 1/[f'(x)*f^-1(10)]

3. The attempt at a solution

a.) a function is monotonic when it is either always increasing or always decreasing. You can check by looking at the derivative, f(x)' = 3x^2 + 4. This Function is one to one because it is monotonic (always increasing).

b.) I simply cannot figure this out. I am pretty sure our professor does NOT want us to try to find a direct equation for f^-1(x). I believe she wants us to use the method she calls "inspection", to look at the problem carefully and figure out a y value, then f^-1(x) = y, so I can find f(10). pretty lost.

c.) See above... though I know the equation is f^-1(x)' = 1/[f'(x)*f^-1(10)]
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 18, 2010

### tiny-tim

Hi srfriggen!

(try using the X2 tag just above the Reply box )
Yes, part a) told you f(x) is increasing, and you can immediately see that f(0) = 6 and f(1) = 11, so f-1(10) must be between 0 and 1 …

now narrow it down further.

3. Feb 18, 2010

### srfriggen

That helps a little but doesn't really get to an answer. I have a feeling she meant to write 3x instead of 4x. That would make things a lot more elegant.

4. Feb 18, 2010

### Dick

She may also have meant f^(-1)(-10) instead. That would make more sense as well.

5. Feb 18, 2010

### srfriggen

True, -10 would work nicely.

ok, I've been going batty trying to figure this one out. It was on a small pop quiz last night and she said it shouldn't have taken us more than a minute to do that problem...

so am I nuts and terrible at calc or does it seem like she made a mistake??? cause I don't see an easy solution to this problem at all.

6. Feb 18, 2010

### Svensken

(x-10)^1/2=f^-1(x)

or am i completely wrong

7. Feb 18, 2010