Understanding 1/f(x) of Graphs with No Equation

[ProPM]

Hi,

I have a graph of a function, but it has no specific equation, just a drawing, denoted by f(x). The question asks me to do the 1/f(x) of that graph. What does that mean?

Thanks in advance.

 

[blather]

The easiest way to get this is to flip the x and y axes around.

So, you could draw an imaginary line (like a dashed line) along the x-y line (45 degrees from the x-axis in quadrant 1 going through the origin).

Then, take the lines under the dashed line, draw them on top like they'd be seen in a mirror. The lines on the top go to the bottom.

 

[Char. Limit]

The easiest way to get this is to flip the x and y axes around.

So, you could draw an imaginary line (like a dashed line) along the x-y line (45 degrees from the x-axis in quadrant 1 going through the origin).

Then, take the lines under the dashed line, draw them on top like they'd be seen in a mirror. The lines on the top go to the bottom.

That, to me, sounds like a method to graph the inverse of f(x). I think what he's looking for, if he hasn't misinterpreted f^-1(x) as 1/f(x), is the reciprocal of f(x).

If you're looking for the inverse, ProPM, the method above will do. If you're looking for the reciprocal, the best method I can think of is to find points of f(x) on the graph you're given and find the reciprocal of those points, and plot them as (x₀, 1/f(x₀)).

 

[ProPM]

Oh, yes, I want the reciprocal. So to find the inverse, I simply do, for my attachment for example, f(x): when x = 2 and y =2, when x = 2, y = 0.5?

And, if you wouldn't mind, there's this question here I need help with:

Show that the equation (k-1)x² + 2x - (k-3) = 0 has real roots for all values of k.

This is what I did:

b²- 4ac > 0
4 - 4(k-1)(-k+3) = 0
4 + 4k² - 16k + 12 = 0
4k² - 16k + 16 = 0

If we solve, we get a real answer, 2 in this case.

But how can I prove it is true for all values of k.

I hope I am not bothering you guys,

Thanks,
Pro PM

 

[Ray Vickson]

Hi,

I have a graph of a function, but it has no specific equation, just a drawing, denoted by f(x). The question asks me to do the 1/f(x) of that graph. What does that mean?

Thanks in advance.

If you can measure (approximately) some of the (x,y) positions on a piece of graph paper, you can get the graph of y = 1/f(x) (approximately). For example, on the y = f(x) curve if at some value of x we have y = 3, then at that same x but on the y = 1/f(x) curve we would have y = 1/3. In other words, if f(x) = 3 then 1/f(x) = 1/3, etc. That's all!

RGV

 

[SammyS]

...
And, if you wouldn't mind, there's this question here I need help with:

Show that the equation (k-1)x² + 2x - (k-3) = 0 has real roots for all values of k.

This is what I did:

b²- 4ac > 0
4 - 4(k-1)(-k+3) = 0
4 + 4k² - 16k + 12 = 0
4k² - 16k + 16 = 0

If we solve, we get a real answer, 2 in this case.

But how can I prove it is true for all values of k.

I hope I am not bothering you guys,

Thanks,
Pro PM

4 - 4(k-1)(-k+3)
= 4( k² -4k +4)
= 4(k-2)²

And what can you say about (k-2)² ?

 

[ProPM]

That no matter if k is inputted as negative or positive, if it is being squared, it will always be outputted as positive?

 

[Char. Limit]

That no matter if k is inputted as negative or positive, if it is being squared, it will always be outputted as positive?

Or zero. But the square root of 0 is defined as well.

 

[ProPM]

Wow, great. Thanks a lot guys! When I grow up I want to be just like you

 

[SammyS]

That no matter if k is inputted as negative or positive, if it is being squared, it will always be outputted as positive?

Isn't (k-2)² ≥ 0 for all k?

You can also say that \sqrt{4 - 4(k-1)(-k+3)} = \sqrt{4(k-2)^2}=2\sqrt{(k-2)^2}=2|k-2| \ge 0 for all k.