Question

In need of some proof assistance.
The function \(f(x)=3x^5- 10x^3+15x+1\) is strictly increasing, and therefore, invertible on R. Find \((f^{-1})'(1)\)

Answer 1

f' = dy/dx

(f^-1)' = 1/(dy/dx) = dx/dy

Answer 2

I tried to use the inverse function thm, but I ended up with 1/0...

By the inverse function thm. since this qualifies due to being strictly monotone and continuous since it is a polynomial, We can use \[(f^{-1})'(y_0)=\frac{1}{f'(x_0)}\]

Answer 3

But my issue is that at x=1 f'(x) =0

Answer 4

f(x) = 3x^5- 10x^3+15x+1

f' = 15x^4 -30x^2 +15

f'(1) = 30-30 = 0 correct

Answer 5

as such the equation of the inverse of f(x) at x=1 is the line x=1 since we have a vertical slope

Answer 6

but I need to find the value of the derivative of the inverse at that point
so I wanted to use this example. I'll type it, but I don't know how

Answer 7

Let n be a positive interger and let f(x)=x^n for x>0. Then f is strictly increasing on (0,\infty) and its invers is the nth root function g(y)=y^{1/n}.

Answer 8

then it just does the computations for g'

Answer 9

So, my issue is, how can I apply this thm.?

Answer 10

and I need the value of the derivative of the inverse at x=1

Answer 11

given y= f(x) = x^n : g(y) = g(f(x)) = x if g = y^(1/n)

Answer 12

that is the example, but i thought we might be able to use it somehow

Answer 13

this simply tells us that the inverse of x^n is x^(1/n)

Answer 14

our f(x) is not strictly increaseing on the interval tho

Answer 15

it is

Answer 16

f'(x) = 0 when x=1 0 isnot increasing

Answer 17

it is, it's called strictly increasing ie \(f(x)\le f(y)\le f(z)\)

Answer 18

if we observe alphabetical order

Answer 19

with a slope of 0, the function remains the same value, but does not decrease

Answer 20

yeahyeah ... f(x) is increasing

Answer 21

x^1/3 is an increasing fucntion yes?

Answer 22

yep

Answer 23

what is the slope at x=0?

Answer 24

0

Answer 25

really?

Answer 26

oh wait, 1/3 not 3

Answer 27

oops, let me look at it

Answer 28

1/3 x^{-2/3} at x=0we get a dne, but lim is 0

Answer 29

there is a vertical slope at x=0

our critical points of interest are when f'=0 or f' is undefinied

Answer 30

well, not technically

Answer 31

since the function would have to be piecewise, for it to be a vertical slope

Answer 32

undefined does not mean a tangent line does not exist, since the most standard form of a line is :

Ax + By = C

Answer 33

otherwise, it could be horizontal

Answer 34

also, I think we may be getting off track, as critical points are not allowed to be used here

Answer 35

the equation of a tangent line that has a vertical slope is simply x = k for some constant k

the slope is not undefined ... it is a vertical slope

Answer 36

we are not allowed to use that. It has not been proven

Answer 37

Literally all I can use is the definition of derivative and the inverse function thm

Answer 38

and the definition of continuous if necessary

Answer 39

a vertical slope is by definition of derivative ... the left and right limit to +inf they have the same limit

Answer 40

we don't have left and right limit

Answer 41

.... youll have to explain to me why we dont

Answer 42

because this is an analysis course and if it has not been explicitly proven we are not allowed to use it

Answer 43

which is evil, considering I know the answer I just can't arrive there with the tools I am given

Answer 44

what is your definition of derivaitve ..

Answer 45

\[L:=lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}\]

Answer 46

where L is then =f'(x)

Answer 47

sorry f'(c) not x

Answer 48

and the limit is defined as when the limit from the right and left are the same .... or what is your definition of a limit?

Answer 49

it's the epsilon delta definition

Answer 50

or the sequential definition, but we are not allowed to use left and right

Answer 51

I've been marked wrong for it before

Answer 52

let c = x+h

\[L:=lim_{x \rightarrow (x+h)} \frac{f(x)-f(x+h)}{x-(x+h)}\]
\[L:=lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\]

Answer 53

We are not allowed to go there. Unfortunately since that is the def I am more familiar with using

Answer 54

write up you delta epsilon efinition for me

Answer 55

If you don't know it, no offense, but I doubt you can do this proof. That is essential pre-requisite knowledge for it.

Answer 56

i ahvent taken a real analysis class perse, but i have a read the old textbooks

Answer 57

yea, unfortunately, (and believe me I wishhhhhh this was not the case) I don't think you can get this particular proof

Answer 58

given e > 0, |x-c| > d something like that

Answer 59

prolly not :)

Answer 60

yea, If for every epsilon greater than zero there exists a delta greater than zero such that the absolute value of x-c is less than delta then the absolute value of f(x)-L is less than epsilon

Answer 61

the absolute value bars in the definition infer a left right limit ... but yeah, without your course material id be at a loss

Answer 62

I tried using left and right limits with continuity and I got yelled at
:/

Answer 63

yell back :)

Answer 64

hence my lack of wanting to use them again. Haha, well maybe if I was better at writing rigorous proofs I could

Answer 65

I ended up just saying that x=1 is implied by g'(y)=1/0

Answer 66

f(x) is invertible, f'(1) = 0 ... therefore the inverse is 1/0 which is undefiined, or vertical depends on how your course defines 1/0

Answer 67

I can't think of how to prove it so hopefully they will just ignore it

Answer 68

The issue is that 1/0 was never defined

Answer 69

wanna try another?

Answer 70

the derivative of f(x) is never negative, so f(x) is increasing on any function

by the thrm, f(x) is invertible on this interval ... etc

Answer 71

*f(x) is increasing on any interval

Answer 72

Got it

Answer 73

here it is essentially: we have (f^{-1})'f(x)<-I equal y!!!=1/f'(x_0)
so when does f(x)=1? at x=0 so our x_0=0 and we get 1/15

Answer 74

thought you'd like to know