Question

Find the derivative

Answer 1

Find (f^-1)'(a)
\[f(x) = x^3 + 2x - 1, a = 2\]

Answer 2

Take that word file. Hope this is helpful for you. I am using MathType for equation typing.

Answer 3

And this is a screen shot if you don't have MathType.
If you need any help, I won't be late for you.

Answer 4

There is nothing in the file except "find the value of a=2"... my internet is wigging out right now, so idk if it's just me?

I was sick when we covered this in class and the answer book doesn't give explanations, so I'm trying to figure out how to do these problems :P

Answer 5

That is exactly what I got! 3x^2 + 1 for the derivative, f'(2) = 14, f(2) = 11... but where do I go from there?

Answer 6

Did you get the pic I sent after the word file?

Answer 7

Yes, thank you! Sorry my internet was being slow.

I don't get what I'm supposed to do next though... would you mind explaining? thanks so much :)

Answer 8

@3mar still there? :)

Answer 9

Yes

Answer 10

Specify the point I can start from it.

Answer 11

I understood everything you did, but the answer book says the answer is 1/5... how do I get to 1/5 from the point you left off?

Answer 12

You noticed we're taking the inverse derivative right?

Answer 13

Ok

Answer 14

I got it

Answer 15

Take that

Answer 16

:D

Answer 17

Figure this one out Scrabble? :D

Answer 18

Okkkk I see what you did wrong little lady.
We're not using the point x=2.
Notice this value is being plugged into `the inverse function`.
So this corresponds to y=2.

Answer 19

You need to start by finding the x-value which corresponds to the given y,\[\large\rm 2=x^3+2x-1\]

Answer 20

Oh! Niceeee zep :D

Answer 21

3 = x^3 + 2x
and then....

Answer 22

No no move the 3 to the right side.
Solving a cubic can be a real pain.

But for this one, there should be a really obvious x value that works.

Answer 23

1 :D

Answer 24

But like... what would I have done if it wasn't so obvious?

Answer 25

If it wasn't so obvious,
you would have to apply your rational root theorem.\[\large\rm 0=qx^3+x~stuff+p\]Dividing the q out of each term,\[\large\rm 0=q\left(x^3+x~stuff+\frac{p}{q}\right)\]If the polynomial has a rational root, then the rational root must have factors coming from this p/q.

So for our problem, it was a simple -3, right?
Factors of -3:
\(\large\rm 1\cdot-3\)
\(\large\rm -1\cdot3\)

So if x=1 had not worked,
we had three other options we could have tried, x=-1, x=3, x=-3.
If none of those had worked, then the problem has no rational roots.
They probably won't give you a problem with no rational roots though.

Answer 26

You can see how this gets really burdensome though.
Like if it hadn't been a -3 but instead a -12 then we get A TON of options.

Answer 27

Have you learned the fancy formula for finding derivative of an inverse function? If not that's ok. We can derive it really quickly using our chain rule.

Answer 28

I was absent but yeah, we're supposed to use that

Answer 29

What is this fancy formula you speak of?

Answer 30

Is it like 1/f'(f^-1x) or something?

Answer 31

Well recall that if you take the composition of a function and it's inverse, you simply get the argument back, right?\[\large\rm f(f^{-1}(x))=x\]You probably remember this back in trig quite a bit, \(\large\rm arccos(cos(x))=x\)
stuff like that.

Taking derivative of both sides, chain rule on the left side of the equation,\[\large\rm f'(f^{-1}(x))\cdot (f^{-1})'(x)=1\]Dividing by the f'(f^-1(x)) gives us our formula,\[\large\rm (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]Yes that's the one :)

Answer 32

\[\large\rm (f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}\]

Answer 33

How would I find f^-1(2) ? I feel like it's obvious but I'm not seeing it xd

Answer 34

Ya it is :3

Answer 35

Lol hold on

Answer 36

Is it 2?

Answer 37

I mean 3

Answer 38

\[\large\rm f(x)=y\qquad\to\qquad f^{-1}(y)=x\]

Answer 39

We started with y=2,
and what x did we end up with?

Answer 40

Oh, 1

Answer 41

\(\large\rm f^{-1}(2)=1\)
Good good good.

Answer 42

Ah ah ah, I see. Then plug that into the derivative, get 1/5, ya ya.
Thanks zeppers! Mind walking me through one more?

Answer 43

Ya that can be tricky huh?
You do all that messy work and it boils down to simply plugging x=1 into the derivative, instead of x=2.

Sure

Answer 44

Yeah I know, one stupid little mistake and the whole problem is wrong :(
Here is the next one:
\[f(x) = sinx, \frac{ -\pi }{ 2 }<x<\frac{ \pi }{ 2 }\]
The signs should be less than or equal to.

Answer 45

Same instructions, find f^-1(a)

Answer 46

So.... 1/2 = sinx ?

Answer 47

What is a?

Answer 48

Oh, a = 1/2
dunno why that didn't copy

Answer 49

Ya let's start there.

Answer 50

x = pi/6 then

Answer 51

And the derivative is cosx.

Answer 52

k

Answer 53

So would the answer be sqrt3/2 ?

Answer 54

Hmm that doesn't seem right.

Answer 55

\[\large\rm (f^{-1})'\left(\frac12\right)=\frac{1}{f'\left(f^{-1}\left(\frac12\right)\right)}\]

Answer 56

I am sorry Abbles for the irrelevant answer, and thanks to zepdrix for this brilliant help. It is useful for myself. Thanks zepdrix.

Answer 57

No prob 3mar and yeah, zep is pretty brilliant :)

Answer 58

Zep, would it be 2/sqrt3?

Answer 59

\[\large\rm (f^{-1})'\left(\frac12\right)=\frac{1}{f'\left(\color{orangered}{f^{-1}\left(\frac12\right)}\right)}\]So then,\[\large\rm (f^{-1})'\left(\frac12\right)=\frac{1}{f'\left(\color{orangered}{\frac{\pi}{6}}\right)}\]

Answer 60

Yesss that looks better!

Answer 61

Haha :D

Answer 62

Or 2sqrt(3)/3 if they want the answer rationalized.

Answer 63

Thanks zepples. Why are you up so late?

Answer 64

Nah, my teacher doesn't care about rationalizing (is that abnormal for calc teachers?)

Answer 65

No, once you get passed Pre-Calc that becomes pretty normal.
There are a lot of weird formulas involving square roots in the bottom,
so it doesn't make a lot of sense to fuss over things like that at this levle.

Answer 66

past? woops

Answer 67

Up so late??
Pshhhh you know how I do!
Up all night, I'll sleep when I die!!
The streets is the only place I call home.
Yehhhhhhhhh

Answer 68

Err *cough* scuse me -_-
not sure where that came from..

Answer 69

Funny because in order to get PAST precalc, you have to PASS it... okay, so not that funny...
Hahaha ZepxD

Answer 70

lol :2

Answer 71

:3*