Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

Use the function to answer the question. f(x) = 5x - 10 What is the value of f^–1 (–3)?

OpenStudy (anonymous):

I got 12

ganeshie8 (ganeshie8):

lets test if its correct.. put x = 12 in f(x), do u get -3 ?

ganeshie8 (ganeshie8):

f(12) = 5*12 - 10 = 60 - 10 = 50 so, 12 is not right ok

OpenStudy (anonymous):

ohh.... ok. OH, wait, sorry, I got 12 on another question. I don't know this one

ganeshie8 (ganeshie8):

to find f^–1 (–3), u need to solve x below :- -3 = 5x - 10 solve x

OpenStudy (anonymous):

Ok, I got 1.4

ganeshie8 (ganeshie8):

x = 7/5 = 1.4 is \(\large \color{red}{\checkmark}\)

OpenStudy (anonymous):

Ok, sweet. Thank you. Now is this 12? Use the functions to answer the question. f(x) = 4x - 7; g(x) = x + 3 What is the value of (f circle g)(4)?

ganeshie8 (ganeshie8):

looks wrong, lets do it again quick ok

ganeshie8 (ganeshie8):

(f o g)(4) f( g(4) ) ^^^

ganeshie8 (ganeshie8):

first find g(4) wat do u get ?

OpenStudy (anonymous):

How would I do that?

ganeshie8 (ganeshie8):

we're given g(x) = x + 3 to find g(4), simply put x = 4 above g(4) = 4+3

ganeshie8 (ganeshie8):

(f o g)(4) f( g(4) ) f (4+3) f(7)

ganeshie8 (ganeshie8):

find f(7) now give it a try :)

OpenStudy (anonymous):

9?

OpenStudy (anonymous):

O_o

ganeshie8 (ganeshie8):

try again, and do show me wat u tried ok :)

OpenStudy (anonymous):

to find g(7), put x = 7 above g(7) = 7+3 And so then 7 + 3 = 10 so g(10)

ganeshie8 (ganeshie8):

we're given, f(x) = 4x - 7 so, to find f(7), simply put x = 7 above :- f(7) = 4*7 - 7 = 28 - 7 = 21

OpenStudy (anonymous):

Oh, I was way off

OpenStudy (anonymous):

10. Use the functions to answer the question. f(x) = 4x – 1; g(x) = 5x2 Which is equal to (f circle g) (x)? (Points : 4)

ganeshie8 (ganeshie8):

no, you're doing good, by next question u wil be on track :)

ganeshie8 (ganeshie8):

this question u wil be doing urself ok

ganeshie8 (ganeshie8):

(f o g)(x) f( g(x) )

ganeshie8 (ganeshie8):

plugin g(x) above

ganeshie8 (ganeshie8):

(f o g)(x) f( g(x) ) f( 5x2 )

ganeshie8 (ganeshie8):

plugin x = 5x2 in f(x)

OpenStudy (anonymous):

Okay

ganeshie8 (ganeshie8):

(f o g)(x) f( g(x) ) f( 5x2 ) 4*5x^2-1 20x^2 - 1

ganeshie8 (ganeshie8):

see if that makes some sense

OpenStudy (anonymous):

Alright, it does! I got it. Now, I I got the answer (-5, -8) on this on 3. Use the function to answer the question. f(x) = -3|x + 5| + 8 What is the vertex of the graph of f(x)?

OpenStudy (anonymous):

Any Ideas?

ganeshie8 (ganeshie8):

f(x) = a|x-h| + k vertex = (h, k)

OpenStudy (anonymous):

So 5 and 8

ganeshie8 (ganeshie8):

try again

OpenStudy (tester97):

great work helping and explaining @ganeshie8 ^_^

ganeshie8 (ganeshie8):

aww ty tester, jonathan is quick learner :) f(x) = -3|x + 5| + 8 = -3|x -(- 5)| + 8

OpenStudy (anonymous):

OH!!! Cause you have to take the - sign, so you get.... Wait..... OH!!!! -5 and 8?

ganeshie8 (ganeshie8):

yes, you got it !

OpenStudy (anonymous):

YAY!!! Lol. 4. Use the function to answer the question. f(x) = |x + 1| What is the range of f(x)? (Points : 4) all real numbers greater than 0 all real numbers greater than or equal to 0 ---> all real numbers greater than or equal to -1 all real numbers greater than -1

ganeshie8 (ganeshie8):

hint : range = all y values domain = all x values

ganeshie8 (ganeshie8):

f(x) = |x + 1| f(x) = |x + 1| + 0 ^^^

ganeshie8 (ganeshie8):

thats the y-coordinate of vertex

OpenStudy (anonymous):

So it would be A

ganeshie8 (ganeshie8):

try again

OpenStudy (anonymous):

Then B?

ganeshie8 (ganeshie8):

yes, it includes 0 also cuz vertex will be there at (-1, 0) so range will >= 0

ganeshie8 (ganeshie8):

*be

OpenStudy (anonymous):

Thank y'all for your help!!!! :D

ganeshie8 (ganeshie8):

np :) u wlc !

OpenStudy (anonymous):

You want to solve the equation -3 = 5x - 10. That is not the solution to this equation.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!