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Suppose that a one-to-one function f has tangent line y = 5x+ 3 at the point (1, 8). Evaluate (f^{-1})'(8)

 Oct 16, 2015

Best Answer 

 #3
avatar+27470 
+5

I thought this was interesting too.  I'd like to make it both simpler (1. below) and more complicated (2. below)!

 

inverse functions

.

 Oct 18, 2015
 #1
avatar+3885 
+5

\(\mbox{what do you know about the relationship between }f^\prime(x) \mbox{ and }(f^{-1})^\prime(x) ?\)

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 Oct 16, 2015
 #2
avatar+97497 
+5

Rom, I hope that you do not mind me butting in but i could not answer your questions straight off and I wanted to think about what was happening myself.

 

So whether or not you are teaching the question asker, you are teaching me:)

 

I developed a formula for a curve that met the given criterion.

 

This is what I came up with.  

 

https://www.desmos.com/calculator/7uhpqmscci

 

 

I/You did not need to do any of this to answer the question, I was just thinking laterally. :)

 

The answer is very simple.

 Oct 17, 2015
edited by Melody  Oct 17, 2015
 #3
avatar+27470 
+5
Best Answer

I thought this was interesting too.  I'd like to make it both simpler (1. below) and more complicated (2. below)!

 

inverse functions

.

Alan Oct 18, 2015
 #4
avatar+97497 
0

Thanks Alan,

 

 

In response to 1).

 

Yes of course I could have   frown      LOL

BUT

The graph would not hve looked as interesting   laugh

 

 

 

In response to 2)

That makes my head hurt.    angry

Maybe I will think about it later after something deadens the pain.      angrycryingangry       LOL   again    wink

 Oct 18, 2015
 #5
avatar+27470 
+5

1. You are right Melody, the graph would have been really boring!  

 

2. Here are a couple of graphs to illustrate the solution when n = 2 (so f(x) = x5 + 7) 

 

n=2

.

 Oct 18, 2015
 #6
avatar+97497 
0

Thanks Alan,  I still like my graph.          laughwinklaugh   

 

I need to look at your number 2 answer properly :))

 Oct 18, 2015
edited by Melody  Oct 18, 2015
 #7
avatar+3885 
0

all very nice work but you all seemed to have missed the point

 

\(\left(f^{-1}\right)^\prime(x)=\dfrac 1 {f^\prime(f^{-1}(x))}\)

 

\(\mbox{So as }f \mbox{ has a tangent line }y=5x+3, @(1,8) \\ {f^{-1}}^\prime(8) = \dfrac 1 {f^\prime(1)}=\dfrac 1 {5(1)+3}=\dfrac 1 8\)

.
 Oct 20, 2015

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