# Thread: Find a differentiable function with the given properties

1. ## Find a differentiable function with the given properties

find all function $\displaystyle f$ differentiable over $\displaystyle \mathbb{R}$
$\displaystyle - (\forall x\in \mathbb{R}) : \big(f'(x)\big)^2-\big(f(x)\big)^2=1 \quad (1)$
$\displaystyle -f'(0)=1 \quad(2)$
$\displaystyle -f'$ is differentiable over $\displaystyle \mathbb{R}\quad (3)$

2. Ok not to be a pain but the word is differentiable not derivable.

Taking the square root

f ' ^2 -f = 1 or -1

Differentiate -- in either case

2f " f ' -f ' = 0

f ' (2f"-1) = 0

Now solve f ' = 0 (use original equation to find values of c) and 2f " - 1 = 0

That should be enough to get you to the end

3. Originally Posted by Calculus26
Ok not to be a pain but the word is differentiable not derivable.

Taking the square root

f ' ^2 -f = 1 or -1
either i'm missing something or you made a mistake here

4. Calculus26 it's $\displaystyle \big(f'(x)\big)^2-\big(f(x)\big)^2=1$ and not $\displaystyle \big(f'(x)\big)^2-f(x)=1$

5. Sorry I read it too quickly I was seeing

[f '^2 -f]^2 = 1 for the original

For (f ')^2 - f^2 =1

the process is still the same

2f ' f "- 2 f f '= 0

f ' = 0 or f '' -f = 0

yields f = c from which there are no solutions

or f " -f = 0 which yield Acos(x) + Bsin(x) now apply f '(0) = 1

and finish

6. I believe there are no solutions as f ' (0) =1 yields f =Acos(x) +sin(x)

Plugging this into the original there are no values of A Solving the equation.

7. Originally Posted by Calculus26
or f " - f = 0 which yield Acos(x) + Bsin(x) now apply f '(0) = 1
sorry but you are wrong , you mean

f " + f = 0 which yield Acos(x) + Bsin(x)

but here we have f " - f = 0

8. Originally Posted by linda2005
sorry but you are wrong , you mean

f " + f = 0 which yield Acos(x) + Bsin(x)

but here we have f " - f = 0
yes, take $\displaystyle f(x) = Ae^x + Be^{-x}$ instead

9. Ok I've made every possible mistake on this to now--thats what i get for hurrying

we get f = Ae^t + Be^-t

f ' (0) yields A+B =1

Substituting into original equation yields -4AB =1

Now solve --solutions are when A = (2^1/2+1)/2 or (1-2^1/2)/2

with corresponding values for B = 1 -A

Sorry for the mistakes

10. Originally Posted by Calculus26
Ok I've made every possible mistake on this to now
i'm sure you can find more mistakes to make if you tried

Sorry for the mistakes
it happens to the best of us, don't worry about it. mistakes were corrected before they caused much damage.

did you double check your solutions for A?

i'm not going to check them, i trust you

11. I rechecked for both A and B in the original both by hand and with Mathcad to make sure-- thanx for the kind words

But still I always told my students to slow down --Calculus is more about patience than intelligence I'd say--too bad I don't follow my own advice.

That's what happens a year removed from teaching--skills are slipping

12. I find $\displaystyle f(x) = \frac{e^x - e^{-x}}{2}$ I'm wrong or you are Calculus26( f ' (0) yields A+B =1 false we have f ' (0) yields A-B =1) thanks you guys xxx

13. Damn--- you're right using the initial condition A - B = 1 not A+B =1

as I said before--my solution satisfies the equation which is where I checked my results but not the initial conditions.

Now I quit

3 mistakes on one problem

14. By the way not that it matters that much but

is also known as the hyperbolic sine and written

= sinh(x)

You may see the answer written that way

Thanx for your kindness amongst my blunders I should have stopped after

describing the process.

15. Originally Posted by Calculus26
Thanx for your kindness amongst my blunders I should have stopped after

describing the process.
i told you you could find more mistakes if you tried

that was mean. i am only messing with you Calc. but yes, if you find you made 2 blunders, just resort to describing the process and letting the OP do the groundwork. this comes with experience on forums

take care