Determine Direct and Inverse Image f(E) and f^-1 (G)

[phillyolly]

Homework Statement

Let f(x):=1/x^2, x not equal 0, x belongs R
a) Determine the direct image f(E) where E:= (x belongs R : 1<=x<=2)
b) Determine the inverse image f^(-1)(G) where G:= (x belongs R : 1<=x<=4)

Homework Equations

The Attempt at a Solution

A) Let f: R -> R be defined by f(x):=1/x^2. Then, the direct image of the set E:=(x:1<=x<=2) is the set f(E)=(y:1<=x<=1/4).
If G:= (y : 1<=x<=4), then the inverse image of G is the set f^-1 (G)=(x:

And here I don't quite understand how to find an inverse image. Please help.

 

[lanedance]

how about starting by finding the inverse map...

if you choose a reasonable domain f will be 1:1, so its inverse will exist such that f^(-1)(f(x)) = x

 

[phillyolly]

I have no idea what you are talking about, I am sorry...

 

[phillyolly]

What I have done:

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

This is the inverse formula. What should I do now?

 

[lanedance]

I prefer to keep the variables differnt for the function & its inverse as below, i find it makes thing easier, so let:

$y = f(x)$ is a function that maps from x to y

for question a), if the domain of x is [1,2], then the y is in the range [1/4,1] as you've found

now for question b) assuming f is 1:1 we can find its inverse function, let's call it g
$x = f^{-1}(y) = g(y)$ which maps y back to x

 

[lanedance]

now with all that in mind, starting with the function
$$y = f(x) = \frac{1}{x^2}$$

you've said the inverse funtion is
$$x = g(y) = f^{-1}(y) = y^2$$

lets check if it actually satisfies the inverse property
$$f^{-1}(f(x)) = g(f(x)) = (f(x))^2 = (\frac{1}{x^2})^2 = \frac{1}{x^4} \neq x$$

so this is not infact the correct inverse function

 

[lanedance]

to find the correct inverse, start with
$$y = \frac{1}{x^2}$$

now solve for x in terms of y, that gives you g

 

[phillyolly]

I am so lost with your last comment. I can only see that this formula is the same as in your previous post.
Please any clues?

 

[phillyolly]

OK, I think I got it. Please check if I am right,

 

[lanedance]

yep that looks good
$$y(x) = \frac{1}{x^2}$$

so rearranging gives
$$x(y) = \sqrt{\frac{1}{y}}$$

 

[HallsofIvy]

Homework Statement

Let f(x):=1/x^2, x not equal 0, x belongs R
a) Determine the direct image f(E) where E:= (x belongs R : 1<=x<=2)
b) Determine the inverse image f^(-1)(G) where G:= (x belongs R : 1<=x<=4)

Homework Equations

The Attempt at a Solution

A) Let f: R -> R be defined by f(x):=1/x^2. Then, the direct image of the set E:=(x:1<=x<=2) is the set f(E)=(y:1<=x<=1/4).

Everyone has been focusing on (B) but let me make an obvious point 1 is not "less than or equal to" 1/4! Also, it make no sense to say "y such that x has some property". What you should have is f(E)= {y: 1/4<= y<= 1}.

If G:= (y : 1<=x<=4), then the inverse image of G is the set f^-1 (G)=(x:

And here I don't quite understand how to find an inverse image. Please help.

 

[phillyolly]

Hi! This is an old posting of mine. I am going back to this discussion. Is it right that for question b the answer is:

f^(-1)(G)={1/2<=x<=1}

? Thanks!

 

[HallsofIvy]

No, it's not. Notice that if x= -1/2, y= 1/(-1/2)^= 4. f is an even function so its graph is symmetric about the y-axis. lanedance made an error when he said the inverse function is given by $x= 1/\sqrt{y}$. In fact, this function does not have an inverse!

In my first year in grad school, I had to present, before the class, a proof involving $f^{-1}(A)$ where A is a set. I did the whole problem assuming that f had an inverse- very embarassing!

The way you should have approached the problem is to look for x such that f(x)= 1/x^2= 1 and f(x)= 1/x^2= 4. The first equation has solutions 1 and -1, the second has solutions -1/2 and 1/2. Further, the derivative of f is -1/x^3 which is always negative between 1 and 2 and positive between -2 and -1. The fact that the function is decreasing on one interval and negative on the other tells you the domain is just the two intervals, $1< x\le 2$ and $-2\le x< -1$.

If, for some x in either of those intervals, the derivative were 0, you would have to consider the possibility that there is a turning point in the interval.

 

[lanedance]

this isn't very fresh in my head, but I'm not sure I understand

Unless I'm missing something f(x):=1/x^2 on 1<=x<=4 is monotone decreasing with f'(x) not zero anywhere so the inverse exist

 

[HallsofIvy]

But it is not x, in the original function, that is between 1 and 4. The problem was to find $f^{-1}(G)$ where G is [1, 4]. If we write the original function $y= 1/x^2$. Then we are looking for x such that $1\le y\le x$.

Yes, the problem was written in terms of $f^{-1}(x)$ with $1\le x\le 4$ but in terms of the original function, y= f(x), it is y that is between 1 and 4.

Is there a reason why LaTeX is not working?

 

[Ritu Das]

The solution of f(E) is not right because 1≤ y≤ 1/4 would imply the set (-∞,1/4] U [1,∞). But, f(1.5) = 0.444 ∉ (-∞,1/4] U [1,∞) whereas 1.5 satisfies x ∈ [1,2]. Therefore f(E) is derived as follows:

f(E)= f{x∈ ℝ : 1≤ x ≤ 2}

Now, 1≤ x ≤ 2 ⇔ 1≥ 1/x ≥ 1/2 ⇔ 1≥ 1/x^2 ≥ 1/4 ⇔ 1≥ y ≥ 1/4

∴ f(E) = {y∈ ℝ : 1/4 ≤ y ≤ 1}