Find the image set under a function

[grain1]

1. f(x) = 1/4 (x - 1)^2 - 9. What is the image set of the function f? Express in interval notation.

g(x) = 1/4 (x - 1)^2 - 9. (1 ≤ x ≥ 7). Specify the domain and image set of the inverse function g^-1, and find it's rule.

I really don't understand what happens in these questions. I have looked at numerous examples, the easy ones look like it's similar to finding the range, and the hard ones i just don't get. Can someone please break it down for me.

 

[lanedance]

the question asks based on the allowable values of x (the domain) what values can the function x take (the image)

for example say x and f are both defined on the real line, for any x, f can never produce values less than 8\frac{3}{4}, why?

 

[SammyS]

1. f(x) = 1/4 (x - 1)^2 - 9. What is the image set of the function f? Express in interval notation.

g(x) = 1/4 (x - 1)^2 - 9. (1 ≤ x ≥ 7). Specify the domain and image set of the inverse function g^-1, and find it's rule.

I really don't understand what happens in these questions. I have looked at numerous examples, the easy ones look like it's similar to finding the range, and the hard ones i just don't get. Can someone please break it down for me.

What does 1 ≤ x ≥ 7 mean ?

 

[HallsofIvy]

I suspect that grain1 meant to say that 1\le x\le 7.

The graph of g(x) is a parabola with vertex at x= 1, y= -9, opening upward. In particular, for x between 1 and 7, the graph is one side of the parabola so it is "one to one" and there is an inverse. You find that inverse by solving the equation (1/4)(x-1)^2- 9= y of x. That's easy- (x- 1)^2= 4(y+ 9) and you can just take the square root of both sides. Knowing that you want x> 1 tells you which sign to use.

 

[grain1]

Thanks Lanedance...

-8 3/4 is the y - intercept, does this mean my image will be (-8 3/4, ∞)?

SammyS ... yes i meant to type 1≤x≤7.

HallsofIvy... i thought the inverse was found by switching the x and y in the oringinal formula i.e y = 1/4 (x-1)^2 - 9 becomes x = 1/4 (y-1)^2 - 9 and then solve for y

I think i am more confused now than i was.

 

[lanedance]

Thanks Lanedance...

-8 3/4 is the y - intercept, does this mean my image will be (-8 3/4, ∞)?

yes that image is correct, however it is just a concidence x=-8 3/4 is the y - intercept, the important part is that it is the minimum of the function

SammyS ... yes i meant to type 1≤x≤7.

HallsofIvy... i thought the inverse was found by switching the x and y in the oringinal formula i.e y = 1/4 (x-1)^2 - 9 becomes x = 1/4 (y-1)^2 - 9 and then solve for y

to find an inverse never change x & y, the just confuses things, if you have
<br /> y= y(x) = f(x)<br />

see if you can solve for x as a function of y which gives you the inverse function
<br /> x= x(y) = f^{-1}(y)<br />

I think i am more confused now than i was.

 

[grain1]

Thank you again Lanedance, you are being very helpful.

I thought the min was at the vertex... i.e. -9?


g(x) = 1/4 (x - 1)^2 - 9 (1 ≤ x ≤ 7)


g^{-1}(y) = 1/4 (y - 1)^2 - 9
x = 1/4 (y - 1)^2 - 9
4x + 36 = (y - 1)^2
√(4x + 36) = y - 1
+2x + 6 + 1 = y
2x + 7 = g^{-1}

Can you tell me what is next?

can i write the above as my inverse and that will give me an image of (9, 21)?