Is x=2 the correct answer for the inverse of f(x) = 6/(2x+3)?

[SolCon]

Greetings to all.

I have 2 simple questions, one is simply a confirming the answer query, and the other is domain related (like topic title says).

1) In this question, we are asked to show that a²+b² is constant for all values of $$\theta$$. We have a and b as:

a=2sin$$\theta$$+cos$$\theta$$
b=2cos$$\theta$$-sin$$\theta$$

My final answer to this question was:

4sin²$$\theta$$+sin²$$\theta$$+4-4sin²$$\theta$$+1-sin²$$\theta$$

which results in simply a 5. Does this satisfy the question's demands?

2) In this question, we were asked to find an expression in terms of x, for f ^-1(x) and find the domain of f ^-1(x).

The original equation is: 6/(2x+3) for x>=0
So, for f ^-1(x), we'll have: x=6/(2y+3)

The final equation: y=1/2[(6/x)-3]
To get the domain, we place f ^-1(x)=0
So, 1/2[(6/x)-3]=0 , which will give us x=2, right?

After this, the domain must now be structured. But I'm having trouble understanding how this was done. The answer to the question is 0<x<=2. But why is it like this and not 0<=x<=2 instead?

Any help with this is appreciated.

 

[eumyang]

2) In this question, we were asked to find an expression in terms of x, for f ^-1(x) and find the domain of f ^-1(x).

The original equation is: 6/(2x+3) for x>=0
So, for f ^-1(x), we'll have: x=6/(2y+3)

The final equation: y=1/2[(6/x)-3]
To get the domain, we place f ^-1(x)=0
So, 1/2[(6/x)-3]=0 , which will give us x=2, right?

After this, the domain must now be structured. But I'm having trouble understanding how this was done. The answer to the question is 0<x<=2. But why is it like this and not 0<=x<=2 instead?

So the domain is restricted to x ≥ 0 in
$$f(x) = \frac{6}{2x + 3}$$

What is the range of f(x)? Remember that if f(x) is a one-to-one function, the domain of f(x) = the range of f^-1(x) and vice versa. You're saying that the domain of f^-1(x) is 0 ≤ x ≤ 2, which means that you're also saying that the range of f(x) is 0 ≤ x ≤ 2. But is that true? Can f(x) ever equal 0?

 

[SolCon]

Thanks for the reply. :)

One thing though. f(x) may never equal 0 but what if it was x>=2 or x>=4 instead of x>=0 as the condition? And we used f^-1(x)=0 and got x=2?

Would we get:

for when x>=2) x<=2
for when x>=4) 2<=x<=4

Is this right?

 

[eumyang]

I forgot to mention this: you stated that f^-1(x) is
$$f^{-1}(x) = \frac{1}{2}\left( \frac{6}{x} - 3\right)$$

I would go ahead and distribute the 1/2 and clean the RHS up a bit:
$$\begin{aligned} f^{-1}(x) &= \frac{3}{x} - \frac{3}{2}\right) \\ &= \frac{6}{2x} - \frac{3x}{2x} \\ &= \frac{6 - 3x}{2x} \end{aligned}$$

Thanks for the reply. :)

One thing though. f(x) may never equal 0 but what if it was x>=2 or x>=4 instead of x>=0 as the condition? And we used f^-1(x)=0 and got x=2?

Would we get:

for when x>=2) x<=2
for when x>=4) 2<=x<=4

Is this right?

No, it isn't. If the domain of f(x) is restricted to x ≥ 2, what would its range be? Seeing the graph, I notice that for all x ≥ 2, f(x) is always decreasing. So if I find f(2), then I would find the maximum value for the range. f(2) = 6/7, so the range of f(x) is 0 < f(x) ≤ 6/7. f(x) as defined is one-to-one, so the domain of f^-1(x) would be 0 < x ≤ 6/7, not x ≤ 2.

If you say that the domain of f^-1(x) is x ≤ 2, then you are saying that the range of f(x) is f(x) ≤ 2. Given the restricted domain of x ≥ 2, will f(x) ever equal 2?

Now, you try to figure out the domain of f^-1(x) if the domain of f(x) is restricted to x ≥ 4. (Hint: the domain of f^-1(x) is not 2 ≤ x ≤ 4. Also, I don't know what you mean by setting f^-1(x) = 0 to find the domain of f^-1(x).)

 

[SolCon]

Alright.

For the above equation, (6-3x)/2x, this is f^-1(x). So for f(x) (to get range) we will have y [or f(x)]=6/(3+2x). Seeing as x>=2, we plug in the value of 2 in x, and that is how you have gotten 6/7 as the domain. 6/(3+2(2)) = 6/3+4 = 6/7. This is right or wrong?

So, using same method for x>=4, we simply plug in 4 in place of x and get 6/11. This is range of f(x) so domain of f^-1(x) 0<x<=6/11. This is right or wrong?

Also, you said something about the graph. Could you tell me how you constructed it (what values you took) and what the graph looks like (image link?). For the x>=2, I was thinking we must take values of x=2,3,4,5 and so on, in case of x>=2. And in case of x>=4 for x we'd take values 4,5,6,7 and so on. These are the values we take to construct the graph?

 

[eumyang]

Your language is somewhat imprecise, but you have the idea. (Like, "that is how you have gotten 6/7 as the domain..." No, 6/7 is not the domain. 6/7 is the maximum value in the domain of f^-1(x) if the domain of f(x) is restricted to x ≥ 2.)

Here are some graphs. In all 3 cases, the blue is y = x, the red is f(x), and the green is f^-1(x). The domain restrictions aren't really what I say they are, but they work for the purposes of the graphs.

In the http://home.comcast.net/~yeongil/images/F_F-1_0.jpg" , the domain of f(x) is restricted to x ≥ 0.
The range of f(x) is 0 < f(x) ≤ 2, which means that the domain of f^-1(x) is 0 < x ≤ 2.

In the http://home.comcast.net/~yeongil/images/F_F-1_2.jpg" , the domain of f(x) is restricted to x ≥ 2.
The range of f(x) is 0 < f(x) ≤ 6/7, which means that the domain of f^-1(x) is 0 < x ≤ 6/7.

In the http://home.comcast.net/~yeongil/images/F_F-1_4.jpg" , the domain of f(x) is restricted to x ≥ 4.
The range of f(x) is 0 < f(x) ≤ 6/11, which means that the domain of f^-1(x) is 0 < x ≤ 6/11.

 

[SolCon]

I apologize for the late response. -_-

However, I just wanted to give my thanks for helping me realize the relation between function and domain and for clearing up the confusion.

Appreciate the effort with the graphs too.