Question

if f varies jointly as q^2 and h, and f=-36 when q=2 and h=3, find f when q=3 and h=4

Answer 1

For f to vary jointly as q^2 and h, you have the following relationship

\[f = k q^2h\]

Answer 2

awesome! that's what i was thinking.. i worked it out and got f=-108.

is that correct?

Answer 3

where k is your constant of variation.

You need this constant value for the other situation where f is not given

Answer 4

is my answer correct?

Answer 5

I'm interested in seeing how you got that too.

Answer 6

Alright, for brevity's sake,

for when f = -36, q = 2, h = 3

\[-36 = k \times 2^2 \times 3 = 12k \rightarrow k = -3\]
Rewrite that first eqn now to become:\[f= -3q^2k\]When q = 3, h = 4\[f = -3 \times 3^2 \times 4 = -3 \times 9 \times 4 \]\[f = -108\]

Answer 7

So, you were right! :)

Answer 8

YES! sorry, i was trying to type that.. and i was failing miserably...

thanks!

Answer 9

sorry, that bit where you have \[f= -3q^2k\]should be

Answer 10

\[f= -3q^2h\]

Answer 11

awesome.

Answer 12

No problem. I understand..

Answer 13

i have one more problem if you don't mind sticking around?

Answer 14

Shoot on! Let's see if we can work it out.

Answer 15

actually 2 more, but i think one of them will be really quick!

Answer 16

given f(x)=-3x+5 and g(x)=4x-1, find (fog)(x)

Answer 17

I don't understand these (fog) and (gof) questions.

Answer 18

Okay, let's walk through it bit by bit

Given:\[f(x) =-3x+5, g(x) = 4x-1\]
Now, the part where it says (fog)(x) also means f(g(x))

Answer 19

Note that x in f(x) was replaced by g(x), so we're substituting x with g(x) in f(x)

Answer 20

ohhhh! that makes sense! I got:

Answer 21

(fog)(x)=12x+2

Answer 22

So\[f(g(x)) = -3(4x-1) + 5 = -12x +3 +5 \rightarrow f(g(x)) = -12x +8\]

Answer 23

Does that make sense?

Answer 24

right! I see where I messed up! the actual equation is f(x)=3x+5.... I accidentally put a negative on here..

Answer 25

That's my bad!

Answer 26

Okay, with that change you get this\[f(g(x)) = 3(4x-1) + 5 = 12x -3 +5 \rightarrow f(g(x)) = 12x +2\]

Answer 27

Hmm, we all make mistakes! :)

Answer 28

alright, one more and i think i'm ready for this test! (:

Answer 29

this is actually probably pretty easy, but i wasn't there during class when she covered it.

find the inverse of the functions:

f(x)= x/4 -3

Answer 30

and i have no idea how to do this, so this is completely new to me!

Answer 31

Now, let's call f(x) y

So,

y = x/4 - 3

solve for x

multiply through by 4 to remove fraction bit

4y = x - 12
x = 4y + 12

Now, we have x. Switch with y, and that's your inverse

y = 4x + 12

\[f^{-1}(x) = y = 4x + 12\]

Answer 32

Is that clear enough?

Answer 33

what is the f^-1

Answer 34

that's the representation of inverse of a function

Answer 35

gotcha!

i think this one is more difficult..:

f(x)=\[\sqrt{8x+7}\]

Answer 36

So, we say like before
\[f(x) = y = \sqrt{8x + 7}\]solve for x\[y = \sqrt{8x + 7}\]square both sides to remove radical\[y^2 = (\sqrt{8x + 7})^2 = 8x + 7\]subtract 7 from both sides\[y^2 - 7 = 8x\]div bothsides by 8\[x = \frac{ y^2 -7 }{ 8 }\]switch x with y\[y = \frac{ x^2 - 7 }{ 8 }\]\[f^{-1}(x)= \frac{ x^2 - 7 }{ 8 }\]

Answer 37

\[f^{-1}(x) = \frac{ 1 }{ 8 } (x^2 -7)\]

Answer 38

thanks for all of your help! I was able to get an 85 on my test that I would've surely failed without it, so thank you!

Answer 39

Oh, great! glad I could help. Yw :)