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Let f(x) and g(x) be functions with domain \((0,\infty)\). Suppose\(f(x)=x^2\)  and the tangent line to f(x) at x=a is perpendicular to the tangent line to g(x) at x=a for all positive real numbers a. Find all possible functions g(x).

 

Hi, I am trying out this problem but I can't seem to know how to do this.

 

I know that f(x)=x^2 so f'(x)=2x.
and for x=a at g(x), the slope is 2a=-1/(2a). 

 

Help is appreciated!

 Nov 3, 2020
 #1
avatar+111600 
+1

Let f(x) and g(x) be functions with domain \((0,\infty)\). Suppose \(f(x)=x^2\)  and the tangent line to f(x) at x=a is perpendicular to the tangent line to g(x) at x=a for all positive real numbers a. Find all possible functions g(x).

 

ok you already have said that the GRADIENT OF THE TANGENT TO F(X) is

 

f'(x)=2x

 

So the gradient of the tangent to g(x) at any given point must be

   \(g'(x)= \frac{-1}{2x}\)

 

So you have to integrate that to get the value of   g(x)

 

Can you do that?

Do you completely understand ?

 

When you get an answer plot f and g on desmos.   Check you understand what happens.

https://www.desmos.com/calculator

 Nov 3, 2020
 #5
avatar+76 
+1

Hi Melody and asinus,

 

Thank you for you help!

I have plotted in f and g onto Desmos but I am unsure what to do now. How should I continue?

yeliah  Nov 3, 2020
 #2
avatar+10581 
+1

Let f(x) and g(x) be functions with domain \((0,\infty) \) . Suppose \(f(x)=x^2 \)   and the tangent line to f(x) at x=a is perpendicular to the tangent line to g(x) at x=a for all positive real numbers a. Find all possible functions g(x).

Sei f (x) und g (x) Funktionen mit Domäne \((0,\infty) \). \(f(x)= x^2 \) und die Tangentenlinie zu f (x) sei bei x = a  für alle positiven reellen Zahlen a senkrecht zur Tangentenlinie zu g (x). Finde alle möglichen Funktionen g (x).

 

Hello yeliah!

 

\(f(x)=x^2\\ f'(x)=2x\)

 

\(g'(x)=-\frac{1}{f'(x)}\\ g'(x)=-\frac{1}{2x}\)

\(g(x)=-\frac{1}{2}\cdot \int x^{-1}\cdot dx\)

\(g(x)=-\frac{1}{2}\cdot ln |x|+C\)

 

\(\{possible\ g(x)\}\in\{(-\frac{1}{2}\cdot ln |\mathbb Q|+\mathbb Q)\}\)  Is this the correct answer?     

[Added by Melody:  No, this isn't right.  Look at my next post for other points]

                     

The tangent line to f(x) at  x=a \(\large ?\) is perpendicular to the tangent line to g(x) at x=a \(\large ?\) for all positive real numbers a. \(\large ?\) Find all possible functions g(x).

I do not understand this text. Is a the same as | x |? Why?

laugh  !

 Nov 3, 2020
edited by asinus  Nov 3, 2020
edited by asinus  Nov 5, 2020
edited by asinus  Nov 5, 2020
edited by asinus  Nov 5, 2020
edited by Melody  Nov 5, 2020
 #3
avatar+111600 
+1

You have made an error asinus, can you find it?

Melody  Nov 3, 2020
 #6
avatar+76 
+1

Hi asinus,

 

Thank you for you help!

I have plotted in f and g onto Desmos but I am unsure what to do now. How should I continue?

yeliah  Nov 3, 2020
 #7
avatar+111600 
+2

What did you get for g ?

 

What did you plot, if you provided me with a link to your plot I could comment a lot better.

 

Note: Asinus's answer is not correct.   (Although it is better than before)

 

\(g'(x)=\frac{-1}{2x}\qquad \text{That is true}\)

 

But what do you get when you integrate that?

Melody  Nov 3, 2020
edited by Melody  Nov 3, 2020
edited by Melody  Nov 3, 2020
 #8
avatar+111600 
+2

Hi asinus,

 

I did not see your edit till a moment ago.

 

your   g'(x) is now correct  but you have not integrated it to get g

Melody  Nov 3, 2020
 #9
avatar+76 
0

I do not know how to share my graph but i plugged in g as -1/4x^2 

yeliah  Nov 4, 2020
 #10
avatar+111600 
+4

That is not going to help.  It has nothing to do with this question. Asinus got confused.

 

You have 

f(x)=x^2   

 

And you know that

\( g'(x)=\frac{-1}{2x} \)

 

NOW can you integrate that to get the value of g(x)???

 

What is the

\(\int \frac{-1}{2x}\;dx\)       ?

 

Normally, if you do not know the answer, you might do this :

\(\int \frac{-1}{2x}\;dx=\frac{-1}{2}\int x^{-1}\;dx = \frac{-1}{2}\int x^{-1+1}\;dx \)

 

But that leaves us with a power of 0 and that definitely is not right!   

So this should remind you that the answer is a natural log !

 

Now can you or Asinus  do this integral and get the function g  ?

\(g(x)=\int \frac{-1}{2x}\;dx\)

Melody  Nov 4, 2020
 #11
avatar+76 
0

Hi Melody, 

sorry it took so long to get back to you. I did more practice and tried integrating g.

I got -1/2 ln|2x|+C. Is that correct?

yeliah  Nov 5, 2020
 #12
avatar+111600 
+3

I just lost a full post on this, I will try again.    angry

 

Hi Asinus and yellah,

 

Let f(x) and g(x) be functions with domain (0,infinity). Suppose f(x)=x^2 and the tangent line to f(x) at x=a is perpendicular to the tangent line to g(x) at x=a for all positive real numbers a. Find all possible functions g(x).

 

\(\int \frac{1}{x}dx=lnx+c \qquad NOT\qquad \int \frac{1}{x}dx=ln|x|+c\)

 

BUT   x must be in the domain of positive real numbers.

 

That is why the question specifically states that it is referring only to real positive values of x,

 

The introduction of the letter 'a' is confusing but all it is saying is that the tangent of f(x) will be perpendicular to the tangent g(x) is true at any specific value of x which is greater than 0.

'a' is used to indicate any specific constant of your choice.

 

So we have

\(f(x)=x^2\\ g(x)=\frac{-1}{2}\;lnx +c\)

 

 

Here is a graph to show this.  You can change the values of a and c with the sliders.

https://www.geogebra.org/classic/szy6anvd

 

Here is just a picture

 

Melody  Nov 5, 2020
 #13
avatar+111600 
+2

 

\(g(x)=\int \frac{-1}{2x}\;dx\\~\\ g(x)=\frac{-1}{2}\int \frac{1}{x}\;dx\\~\\ g(x)=\frac{-1}{2}*lnx+c\\ g(x)=\frac{-lnx}{2}+c \)

Melody  Nov 5, 2020
 #15
avatar+76 
+2

Hi Melody,

 

Thank you so much for helping me. I under stand so much now!

yeliah  Nov 6, 2020
 #14
avatar+10581 
+1

"I will try again. angry"   

Forgive me Melody, you know my name.
The "a" confused me a bit.
Thank you very much for your detailed description of how to solve this question.

laugh  !

 Nov 6, 2020
 #16
avatar+111600 
+1

Hi yelliah and Asinus,

 

You are both very welcome.

I have improved the graph just a little.

If you need me to explain something in particular then just ask, I will do my best to help you further.

 

And yes Asinus, I do know your name but I do not usually use real names publically unless I am invited to do so. 

(Or unless I am being very absent minded   frown  )

Usernames are for public use.   laugh

 Nov 6, 2020

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