Can you find the value of $f'(20\pi)$ ?

Calculus Level 3

Let $f(x)$ be a function defined for all $x\in\mathbb{R}$ such that $f(x+y)=f(x)\cdot f(y)$ for all $x,y\in\mathbb{R}$ . Let $f'(0)=18$ and $f(20\pi)=\tfrac12$ . What is the value of $f'(20\pi)$ ?

The answer is 9.

1 solution

حكيم الفيلسوف الضائع
May 25, 2014

By the definition of the derivation:

$\eqalign{ f'(x)&=\lim\limits_{h\to0}\dfrac {f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac {f(x+h)-f(x+\color{#D61F06}0)}{h} \\ &=\lim\limits_{h\to0}\dfrac {f(x)f(h)-f(x)f(0)}{h} \\ &=\lim\limits_{h\to0}f(x)\dfrac {f(h)-f(0)}{h-0}=f(x)\underbrace{\lim\limits_{h\to0}\dfrac {f(h)-f(0)}{h-0}}_{\text{definition of the derivative at }0} \\ &=f(x)\cdot f'(0)=18\cdot f(x). }$

Plugging in $x=20\pi$ we get that:

$f'(20\pi)=18\cdot f(20\pi)=18\tfrac12=\boxed9.$

For completeness, you should demonstrate that there exists at least one function which satisfies the conditions.

Note: I would prefer that you explicitly state that $f$ is differentiable at 0, though that is implicitly assumed in stating that $f' (0) = 18$ .

Calvin Lin Staff - 7 years ago

That is precisely the issue. The only differentiable functions that satisfy $f(x + y) = f(x) f(y)$ are $f(x) = 0$ and $f(x) = c^x$ , where $c$ is a positive constant. (In fact, these are the only continuous functions.) And none of them satisfy the given conditions.

Jon Haussmann - 6 years, 7 months ago

Can you find the value of f ′ ( 20 π ) f'(20\pi) f ′ ( 2 0 π ) ?

The answer is 9.

1 solution

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Can you find the value of $f'(20\pi)$ ?