Question

@jim_thompson5910

Answer 1

I have to do parts b and c again, since I did a right.

Answer 2

I agree. Part a) looks perfect. Now I'm checking b

Answer 3

Well I know I did something to get points off since I got -1 and not -2

Answer 4

for part a) you mean?

Answer 5

No b

Answer 6

a is worth 3pts b is worth 2pts and c is worth 4pts

Answer 7

ok one sec

Answer 8

So part b lol

Answer 9

you have this right so far

\[\Large r(x) = \frac{1}{\sqrt{g(2x)}}\]

\[\Large r(x) = \frac{1}{\sqrt{u}}\]

\[\Large r(x) = \frac{1}{u^{1/2}}\]

\[\Large r(x) = u^{-1/2}\]

\[\Large r^{\prime}(x) = \frac{-1}{2}u^{-3/2}*u^{\prime}(x)\]

\[\Large r^{\prime}(x) = \frac{-1}{2}u^{-3/2}*g^{\prime}(2x)\]

Answer 10

so effectively we have

\[\Large r^{\prime}(x) = \frac{-1}{2}(g(2x))^{-3/2}*g^{\prime}(2x)\]

Answer 11

as for the rest of your work in part b), I'm not sure where you're getting that

Answer 12

Because x = 2 so we use where x= 2

Answer 13

and 2*2 = 4

Answer 14

but I would just plug x = 2 into r ' (x) and evaluate

\[\Large r^{\prime}(x) = \frac{-1}{2}(g(2x))^{-3/2}*g^{\prime}(2x)\]

\[\Large r^{\prime}(2) = \frac{-1}{2}(g(2*2))^{-3/2}*g^{\prime}(2*2)\]

\[\Large r^{\prime}(2) = \frac{-1}{2}(g(4))^{-3/2}*g^{\prime}(4)\]

\[\Large r^{\prime}(2) = \frac{-1}{2}(9)^{-3/2}*3\]

\[\Large r^{\prime}(2) = -\frac{1}{18}\]

Answer 15

I see that, but you somehow broke it up but didn't put it back together? I'm not sure exactly what happened

Answer 16

yeah but what about the ^3/2

Answer 17

what do you mean?

Answer 18

You didn't do anything with the ^3/2

Answer 19

you mean when I jumped from

\[\Large r^{\prime}(2) = \frac{-1}{2}(9)^{-3/2}*3\]

to

\[\Large r^{\prime}(2) = -\frac{1}{18}\]

??

Answer 20

Yes I don't see where you used it

Answer 21

\[\Large r^{\prime}(2) = \frac{-1}{2}(9)^{-3/2}*3\]

\[\Large r^{\prime}(2) = \frac{-1}{2}*\frac{1}{9^{3/2}}*3\]

\[\Large r^{\prime}(2) = \frac{-1}{2}*\frac{1}{(9^{1/2})^3}*3\]

\[\Large r^{\prime}(2) = \frac{-1}{2}*\frac{1}{(\sqrt{9})^3}*3\]

\[\Large r^{\prime}(2) = \frac{-1}{2}*\frac{1}{(3)^3}*3\]

\[\Large r^{\prime}(2) = \frac{-1}{2}*\frac{1}{27}*3\]

\[\Large r^{\prime}(2) = \frac{-1}{2}*\frac{1}{27}*\frac{3}{1}\]

\[\Large r^{\prime}(2) = \frac{-1*1*3}{2*27*1}\]

\[\Large r^{\prime}(2) = \frac{-3}{54}\]

\[\Large r^{\prime}(2) = -\frac{1}{18}\]

Answer 22

you probably don't have to go into that much detail when it comes to the steps

Answer 23

I have that answer....

Answer 24

I have on my paper r'(2)=-1/18...

Answer 25

So I don't see why I got a point off then.

Answer 26

hmm maybe the teacher didn't like your approach? not sure

Answer 27

He wouldn't do that he like goes surely off of answer usually idk I'll rewrite it ans approach him tomorrow.

Answer 28

good idea

Answer 29

you have the right final answer, it's just not 100% clear on how you get that

Answer 30

so that might be his issue

Answer 31

Eh it was clear to me lol

Anyways, in part c, I got up to the point and I know all that stuff is right (that's why I got -3 and not -4) but I didn't know how to finish off the problem.

Answer 32

well a lot of the times it all comes down to presentation and communication

Answer 33

time*

Answer 34

I wrote equations thinking I'd have a light bulb go off, but I didn't so ignore those lol

Answer 35

c) Find the values of 'a' if the tangent lines to f(g(x)) and g(f(x)) are perpendicular at x = 3

Answer 36

if we had 2 lines that were perpendicular, what can we say about their slopes?

Answer 37

They are negative reciprocals of each other,

Answer 38

correct, a somewhat equivalent statement is: "the slopes multiply to -1"

Answer 39

so that means

p ' (3) * q ' (3) = -1

where...

p(x) = f(g(x))
q(x) = g(f(x))

Answer 40

your first step is to find p ' (x)

ie derive f(g(x)) with respect to x

Answer 41

It's f'(2a^2)(1/2)

Answer 42

Or aka f'(g(x))(g'(x)

Answer 43

you got f ' ( g(x) ) * g ' (x) which is correct

then you plug in x = 3

f ' ( g(x) ) * g ' (x)

f ' ( g(3) ) * g ' (3)

f ' ( 4 ) * (1/2)

(2a^2) * (1/2)

a^2

Answer 44

And the other is g'(1/a)(5)

Answer 45

be careful...saying f'(2a^2) is NOT correct

Answer 46

saying f ' (4) is the same as 2a^2 is correct

Answer 47

Oh I know how to solve for a. Gotcha.

Answer 48

g( f(x) )

g ' ( f(x) ) * f ' (x)

g ' ( f(3) ) * f ' (3)

g ' ( 1 ) * 5

(1/a)*5

5/a

Answer 49

again, be careful: g'(1/a) is incorrect

Answer 50

Yeah I realized that now lol

Answer 51

Idk why I did that. I was under pressure there was like 5 mins left.

Answer 52

anyways, after deriving both f(g(x)) and g(f(x)), we get the following

a^2 & 5/a

Answer 53

a=-1/5

Answer 54

in order for the tangent lines to be perpendicular at x = 3, the would have to multiply to -1

so,

(a^2)*(5/a) = -1

Answer 55

a = -1/5 is correct

Answer 56

Yeah I know lol

Answer 57

Oh?

Answer 58

hmm? what's your question?

Answer 59

a^2(5/a) =-1 is 5a^2/a the two as divide into each other leaving 5a=-1?

Answer 60

correct

Answer 61

So a= -1/5?

Answer 62

also correct

Answer 63

You just told me it was wrong?

Answer 64

OH i thought you said it was incorrect lol im crazy.

Answer 65

lol that's fine, I was wondering where the issue was