1. ## 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...

2. 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 $\displaystyle y^2+x^2 = R^2$ where R is the radius

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.

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 $\displaystyle f[g(x)]=x$

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

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

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

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

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...

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

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

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

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

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

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

Can you finish it from here?

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

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

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

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

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

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

Can you finish it from here?

3. 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

4. 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

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.

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

$\displaystyle x^2 =3e^2$

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

Done...

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

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

$\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

5. i got range as y >= 2 and less than or equal to 4.5?

Also, for number 10, what are the restrictions?