# Some problems

• Aug 31st 2009, 02:48 PM
ruthvik
Some problems
Find the domain and range of the function

y = 5 - square root of(9 - x^2)

i got the domain
But for the range, my work took a lot of space and came out to something weird. Can anyone work through this and show me how to get the range?

10. Let f(x) = square root of (3 - x).
Find an expression for f inverse x
I got that as y = 3 - x^2

But it says(Be sure to state any necessary domain restrictions???)

11. f(x) = cube root of (x + 2) and g(x) = xcubed - 2

True or false

a. g(x) = f inverse of x
i got as True
b. f of g of x = 1
i got as True
c. The function is one to one
Dont know how to prove this other than graphing and using vert line test
Is there easier way?

20. Express y as a function of x, given that the constant C is a positive number : e ^(y + C) = x^2 + 4

No idea ....lol never did this in last year class

23. Solve each equation without using a calc

a. 2lnx - ln3 = 2
b. ln(x - 2) + ln(x + 5) = 2ln3

For these two, i am not remembering something because these should be easy...im not seeing something here...
• Aug 31st 2009, 04:04 PM
pickslides
Quote:

Originally Posted by ruthvik
Find the domain and range of the function

y = 5 - square root of(9 - x^2)

i got the domain
But for the range, my work took a lot of space and came out to something weird. Can anyone work through this and show me how to get the range?

Do you know what shape this function has? I know that would help me find the range.

Spoiler:
A circle has the form $y^2+x^2 = R^2$ where R is the radius

Quote:

Originally Posted by ruthvik

10. Let f(x) = square root of (3 - x).
Find an expression for f inverse x
I got that as y = 3 - x^2

But it says(Be sure to state any necessary domain restrictions???)

The inverse is correct, you need to restrict the domain of your function so it is 1 to 1.

Quote:

Originally Posted by ruthvik
11. f(x) = cube root of (x + 2) and g(x) = xcubed - 2

True or false

a. g(x) = f inverse of x
i got as True
b. f of g of x = 1
i got as True
c. The function is one to one
Dont know how to prove this other than graphing and using vert line test
Is there easier way?

I agree with a)

for b) I think $f[g(x)]=x$

c) Both functions are 1 to 1. Vertical line test is always a winner! Easier way is to learn the shape of these functions.

Quote:

Originally Posted by ruthvik

20. Express y as a function of x, given that the constant C is a positive number : e ^(y + C) = x^2 + 4

No idea ....lol never did this in last year class

$e ^{y + C} = x^2 + 4$

$y + C = \ln(x^2 + 4)$

$y = \ln(x^2 + 4)-C$

Quote:

Originally Posted by ruthvik
23. Solve each equation without using a calc

a. 2lnx - ln3 = 2
b. ln(x - 2) + ln(x + 5) = 2ln3

For these two, i am not remembering something because these should be easy...im not seeing something here...

$2\ln(x) - \ln(3) = 2$

$\ln(x^2) - \ln(3) = 2$

$\ln(\dfrac{x^2}{3}) = 2$

$\ln(\dfrac{x^2}{3}) = 2\ln(e)$

$\ln(\dfrac{x^2}{3}) = \ln(e^2)$

$\dfrac{x^2}{3} =e^2$

Can you finish it from here?

$\ln(x - 2) + \ln(x + 5) = 2\ln(3)$

$\ln((x - 2) (x + 5)) = 2\ln(3)$

$\ln((x - 2) (x + 5)) = \ln(3^2)$

$(x - 2) (x + 5) = 3^2$

$x^2+3x-10 = 9$

$x^2+3x-19 = 0$

Can you finish it from here?
• Aug 31st 2009, 04:08 PM
ruthvik
So the first one is half a circle?
looks like a circle almost when i graph it but is restricted

For number 10, what would the restrictions be? how did u arrive at yours? graph it?

For the last two

i got up to those parts

wasnt sure of b because it didnt factor but ill just find in decimals

and no idea how to do 23. a. even with that

it says no calc

I agree with everything else u did though, thanks a lot
• Aug 31st 2009, 04:23 PM
pickslides
Quote:

Originally Posted by ruthvik
So the first one is half a circle?
looks like a circle almost when i graph it but is restricted

The negative in front of the square restricts it to a semicircle

Quote:

Originally Posted by ruthvik

For number 10, what would the restrictions be? how did u arrive at yours? graph it?

I found the inverse by swapping x and y, then solving for y.

Quote:

Originally Posted by ruthvik

For the last two

i got up to those parts

wasnt sure of b because it didnt factor but ill just find in decimals

and no idea how to do 23. a. even with that

it says no calc

$\dfrac{x^2}{3} =e^2$

$x^2 =3e^2$

$x =\sqrt{3e^2}$

Done...

$x^2+3x-19 = 0$

For $ax^2+bx+c=0$

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Sep 1st 2009, 10:11 AM
ruthvik
i got range as y >= 2 and less than or equal to 4.5?

Also, for number 10, what are the restrictions?