One method solves the question, the other shows question to be incorrect

Question

Question:

If $f(x+y)=f(x)\cdot f(y) \text{ } \forall\text{ } x, y\text{ } \in R \text{ and } f(5) = 2, f'(0) = 3; \text{ find } f'(5)$

My question:

We know that such types of functions are the exponential functions $f(x) = a^x$. So, $f'(x) = x\cdot a^{x-1}$. So, $f'(0)$ should be $0\cdot a^{-1}$, but it is given $3$.

I don't want the solution to the question given. I want the explanation behind why my working seems to be producing weird results.

UPDATE: Since I posted this question, I realized there's two methods to solve this question. One clearly shown by Emilio below, the other that my textbook does (which does NOT use the fact that f(x) is exponential fn.). Now, interestingly enough, the answer posted below clearly marks the question wrong, while my textbook elegantly solves it.

So, my new question is:

Why those two methods diverge so much, why don't we get the same result for the same question by applying either of them?

My textbook's method for reference:

Let $x=0, y=5$. Putting in given equation implies $f(0) =1$.

Now, $$f'(5)=\lim_{h\to 0}\frac{f(5+h)-f(5)}{h}=\lim_{h\to 0}\frac{f(5)f(h)-f(5)}{h}=f(5)\lim_{h\to 0}\frac{f(h)-1}{h}=f(5)\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=f(5)\cdot f'(0)=2*3=6$$

Final update: So, as stated in comments by dxiv, my textbook solves the problem, but doesn't prove if the function exists in the first place. That's the first time I saw such a thing. I've accepted the current answer because it shows why the question is wrong. Thanks everyone!

Oh, thank you both, I haven't been taught these forms of differentiation. Thank you! — Gaurang Tandon, Dec 18 '16 at 12:26
@GaurangTandon Rewrite $a^x = e^{x\ln a} = g(x\ln a)$ for $g=\exp$. Differentiating with respect to $x$, with the chain rule, you get $f'(x) = \frac{d}{dx}(x\ln a)\cdot g'(x\ln a) = \ln a \cdot e^{x\ln a} = (\ln a) a^x$. — Clement C., Dec 18 '16 at 12:30
Just a comment, but I assume you're aware you don't need to use the result that such an $f$ is an exponential function in order to answer this question. — Dan Rust, Dec 18 '16 at 12:55
Are you sure of the problem? From $f'(0)=3$ we find $f(x)=e^{3x}$ so $f(5)\ne 2$. — Emilio Novati, Dec 18 '16 at 13:08
@EmilioNovati f(x) is an exponential function in a, not in e. I am unsure how you got $f(x)=e^{3x}$ :/ — Gaurang Tandon, Dec 19 '16 at 16:11
my textbook elegantly solves it The collective crystal ball will surely rush to second-guess what that untold method is which your textbook uses. — dxiv, Mar 24 '17 at 03:19
@GaurangTandon The textbook "method" assumes, but doesn't prove, that a function exists which satisfies the given constraints. Compare for example to the classic V.I. Arnold says Russian students can't solve this problem, but American students can -- why?. — dxiv, Mar 24 '17 at 03:43

score 2 · Accepted Answer · answered Dec 19 '16 at 16:26

2

As a first step, as noted in the comments, you have a mistake in the derivative. The function $y=a^x$, can be written as $y=(e^{\ln a})^x=e^{x \ln a} $ , so the derivative ( using the chain rule) is $y'=e^{x \ln a}\cdot \ln a =a^x\ln a$.

The condition $y'(0)=3$ gives $\ln a= 3$ so the function is $y=e^{3x}$ and , for $x=5$, this gives $y=e^{15}\ne 2$.

answered Dec 19 '16 at 16:26

Emilio Novati

64,377

Does that mean the question is wrong? – Gaurang Tandon Dec 21 '16 at 10:55
Yes, there is something wrong in the question. – Emilio Novati Dec 21 '16 at 13:32

One method solves the question, the other shows question to be incorrect

1 Answers1