The derivative of function under constraint

[Raghav Gupta]

Homework Statement

If f(x+y) = f(x)f(y) for all x and y and f(5) = 2, f'(0)=3, then f'(5) is
a) 5
b) 6
C) 0
d) None of these

Homework Equations

Don't know which equation to apply. I was thinking of Rolle's Theorem and mean value theorem here but it is not helping.

The Attempt at a Solution

Found f(0) = 1

 

[pasmith]

Homework Statement

If f(x+y) = f(x)f(y) for all x and y and f(5) = 2, f'(0)=3, then f'(5) is
a) 5
b) 6
C) 0
d) None of these

Homework Equations

Don't know which equation to apply. I was thinking of Rolle's Theorem and mean value theorem here but it is not helping.

You're not actually told that the function is differentiable anywhere other than at zero, so the first thing to do is prove that $f'(5)$ actually exists. Therefore you should start from $$f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}.$$

 

[HallsofIvy]

Notice also that, because f(x+ y)= f(x)f(y), f(x+ 0)= f(x)= f(x)f(0) so we must have f(0)= 1. That helps!

 

[AdityaDev]

Is the answer 6?

 

[SammyS]

Is the answer 6?

I think we're waiting for OP to respond.

 

[Raghav Gupta]

Notice also that, because f(x+ y)= f(x)f(y), f(x+ 0)= f(x)= f(x)f(0) so we must have f(0)= 1. That helps!

But I have already shown that in my attempt of post 1. How it is helping me?

$$f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}.$$

Now what should I do?
I am getting a 0/0 form on right side and L Hopital rule also is not beneficial here.

 

[Dick]

But I have already shown that in my attempt of post 1. How it is helping me?

Now what should I do?
I am getting a 0/0 form on right side and L Hopital rule also is not beneficial here.

Derivative limits always have a 0/0 form. Use f(x+y)=f(x)f(y).

 

[Raghav Gupta]

Derivative limits always have a 0/0 form. Use f(x+y)=f(x)f(y).

So, I am getting,
$$f'(5) = \lim_{h \to 0} \frac{f(5)f( h) - f(5)}{h}.$$
Now f(5) = 2 and f(0)=1, so f(h) = 1
Again getting 0/0 form. What I am supposed to do here ?

 

[Dick]

So, I am getting,
$$f'(5) = \lim_{h \to 0} \frac{f(5)f( h) - f(5)}{h}.$$
Now f(5) = 2 and f(0)=1, so f(h) = 1
Again getting 0/0 form. What I am supposed to do here ?

You are given f'(0)=3. What's that as a difference quotient? Halls gave you a hint in post 3.

 

[Raghav Gupta]

$$f'(0) = \lim_{h \to 0} \frac{f(0)f( h) - f(0)}{h}.$$

Now what? Again 0/0 form.

 

[Dick]

$$f'(0) = \lim_{h \to 0} \frac{f(0)f( h) - f(0)}{h}.$$

Now what? Again 0/0 form.

Think about it some more. The limit is 3. You are given that. As for what f(0) is you were also hinted that.

 

[Raghav Gupta]

Okay , applying L Hopital rule getting f'(h) = 3
Then using in LH rule once again in f'(5) getting 2f'(h) = 6= f'(5)
Was it supposed to be done by LH rule?

 

[Dick]

No, you don't need LH. And what is f'(h)=3 supposed to mean? f'(0)=3. You haven't thought about this enough.

 

[Raghav Gupta]

No, you don't need LH. And what is f'(h)=3 supposed to mean? f'(0)=3. You haven't thought about this enough.

Not,getting
$$f'(0) =3= \lim_{h \to 0} \frac{f( h) - 1}{h}.$$

 

[Dick]

Not,getting
$$f'(0) =3= \lim_{h \to 0} \frac{f( h) - 1}{h}.$$

Good so far. Now figure out f'(5). It has that expression in it. Don't you see?

 

[Raghav Gupta]

Good so far. Now figure out f'(5). It has that expression in it. Don't you see?

Yeah f'(5) = 2*3 = 6 by that expression. Thanks Dick.
Thanks to other people also.

 

[pasmith]

There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.

 

[Raghav Gupta]

There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.

How f(x) = f(1)^x ?
A counter example-
Putting x = 0
We get f(0) = f(1)
Putting x= 5
We get f(5) = f(1)
But f(5) ≠ f(0)
As we are given f(5) = 2 and we know f(0) = 1

 

[SammyS]

How f(x) = f(1)x ?
A counter example-
Putting x = 0
We get f(0) = f(1)

No, f(0) = (f(1))⁰ = 1

Putting x= 5
We get f(5) = f(1)
But f(5) ≠ f(0)
As we are given f(5) = 2 and we know f(0) = 1

 

[Raghav Gupta]

Oh, so it is

No, f(0) = (f(1))⁰ = 1

(f(1))^x not f(1)^x
How can we prove f(x) = (f(1))^x ?

 

[SammyS]

Oh, so it is

(f(1))^x not f(1)^x
How can we prove f(x) = (f(1))^x ?

Ask Mr. (or Ms.) Smith to be certain.

...but

If f(x) = C^x for some C > 0, then we have f(x+y) = f(x)f(y) .

 

[Raghav Gupta]

...but

If f(x) = C^x for some C > 0, then we have f(x+y) = f(x)f(y) .

How? Is it some theorem and if it is what it is called?
Can you tell me that Mr. Sammy ?

 

[SammyS]

How? Is it some theorem and if it is what it is called?
Can you tell me that Mr. Sammy ?

Try it. Use some basic algebra.

If I write it out for you, then Mark and other mentors & homework helpers would be displeased with me. (In fact I would then be displeased with me. )

 

[Raghav Gupta]

Try it. Use some basic algebra.

If I write it out for you, then Mark and other mentors & homework helpers would be displeased with me. (In fact I would then be displeased with me. )

Got it, it was basic algebra really.

There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.

Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))^x
Differentiating,
f'(x) = (f(0))^xlog_ef(0)
Putting x = 0
f'(0) = 0, which is not true since in question it is given f'(0) = 3

 

[SammyS]

Got it, it was basic algebra really.

Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))^x
Differentiating,
f'(x) = (f(0))^xlog_ef(0)
Putting x = 0
f'(0) = 0, which is not true since in question it is given f'(0) = 3

As pasmith said, the conditions given are inconsistent.

Therefore there's an inherent contradiction.

 

[Raghav Gupta]

As pasmith said, the conditions given are inconsistent.

Therefore there's an inherent contradiction.

So following this method the answer could be none of these?
But yes there is error in question as well.
(It is looking like my every question of test that I am asking today is resulting in an error except that matrix one)

 

[pasmith]

Got it, it was basic algebra really.

Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))^x

No, it's $f(x) = (f(1))^x$. As you have established, we must have $f(0) = 1$.

If $f(x + y) = f(x)f(y)$ then $f'(x) = f(x)f'(0)$ and $f(0) = 1$. This initial value problem has the solution $f(x) = e^{xf'(0)}$, so $f(1) = e^{f'(0)}$ and thus $f(x) = (f(1))^x$.

The inconsistency is that if $f'(0) = 3$ then $f(1) = e^3$, whilst if $f(5) = 2$ then $f(1) = 2^{1/5}$.

The question setter has obviously just exploited the fact that if $f(x + y) = f(x)f(y)$ then $f'(x) = f(x)f'(0)$ and has provided arbitrary values for $f(5)$ and $f'(0)$ without realizing that these values are not independent.

 

[SammyS]

If $f(x + y) = f(x)f(y)$ then $f'(x) = f(x)f'(0)$ and $f(0) = 1$.

I think you have the prime in the wrong place.

If $\displaystyle f(x + y) = f(x)f(y)$ then $f'(x) = f'(x)f(0)$ and $f(0) = 1$

Added in Edit:
Never mind. Ray Vickson has pointed out how you were using this.

( I misinterpreted the and in the above .)

 

[Ray Vickson]

I think you have the prime in the wrong place.

If $\displaystyle f(x + y) = f(x)f(y)$ then $f'(x) = f'(x)f(0)$ and $f(0) = 1$

The functional equation gives $f(0) = f(0+0) = f(0)^2$, so $f(0) = 0$ or $f(0) = 1$. We must have $f(0) = 1$, because the other case $f(0) = 0$ would give $f(x) = f(x+0) = f(x) f(0) = 0$ for all $x$, and that would contradict the nonzero value given to $f'(0)$.

We thus have
$$\frac{f(x+h) - f(x)}{h} = \frac{ f(x) f(h) - f(x)}{h} = f(x) \frac{f(h)-1}{h} \\<br /> = f(x) \frac{f(h) - f(0)}{h}$$
Now take $h \to 0$.

 

[Raghav Gupta]

No, it's $f(x) = (f(1))^x$.

If f(x) = C^x for some C > 0, then we have f(x+y) = f(x)f(y) .

Why f(x) is only equal to f((1))^x?
As Sammy is saying it could be any constant.
So we can also write f(x) =(f(2))^x ?

The functional equation gives $f(0) = f(0+0) = f(0)^2$, so $f(0) = 0$ or $f(0) = 1$. We must have $f(0) = 1$, because the other case $f(0) = 0$ would give $f(x) = f(x+0) = f(x) f(0) = 0$ for all $x$, and that would contradict the nonzero value given to $f'(0)$.

We thus have
$$\frac{f(x+h) - f(x)}{h} = \frac{ f(x) f(h) - f(x)}{h} = f(x) \frac{f(h)-1}{h} \\<br /> = f(x) \frac{f(h) - f(0)}{h}$$
Now take $h \to 0$.

It's looking like I'm going round and round if you see my encounter with Dick in this thread.
So if we take
$h\to 0$
Then $$f'(x)= f(x) \frac{f(h) - f(0)}{h}$$
Now what to do?