The Basics

Calculus Level 4

If $2f(x)+f'(x)\leq 8$

for all real numbers $x$ and $f(0)=0$ , find the largest possible value of $f(\ln{2})$ .

The answer is 3.

1 solution

Andreas Wendler
Jan 12, 2016

First we solve the differential equation with inhomogenity I and initial condition f(0)=0 by known entry $f=e^{\lambda x}$ to get: $f(x)=\frac{I}{2}(1-e^{-2x})$ Note that I must be less or equal than 8. So we have: $f(ln 2)=\frac{I}{2}*\frac{3}{4}<=\frac{8}{2}*\frac{3}{4}=3$

Yes, that's the idea. I think you still need to explain why $f(\ln{2})\leq 3$ for any function with $2f(x)+f'(x)\leq 8$ .

Otto Bretscher - 5 years, 5 months ago

Proof is trivial for I<=8. There are not other functions than the ones determined by theory!

Andreas Wendler - 5 years, 5 months ago

Oh but there are ;) You are only finding the functions where $2f(x)+f'(x)$ is a constant less than 8, but $2f(x)+f'(x)$ could be a (non-constant) function whose value stays below 8 for all $x$ .

I'm sorry to play the "advocatus diaboli" once again.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Nothing was said about a completety unknown function as inhomogenity so that such a function can not be included!!!

Andreas Wendler - 5 years, 5 months ago

@Andreas Wendler – Let's say you take a function $f(x)$ such that $2f(x)+f'(x)=8\sin(x)$ . You need to show that $f(\ln(2))\leq 3$ , but your solution does not address a case like this. You only consider the case where $2f(x)+f'(x)$ is a constant $\leq 8$ .

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Call inhomogenity g. Then solution is $f=e^{-2x}F(x)$ with $F'(x)=e^{2x}*g$ and F(0)=0. Following: $f(ln2)=\frac{1}{4}F(ln2)<=8\frac{1}{4}Integral_{0}^{ln2}e^{2x}dx=3$ .

Rem.: Sir, please help me to get a much more detailed formatting guide ( e.g. "Integral" :-((( ) !

Andreas Wendler - 5 years, 5 months ago

The Basics

The answer is 3.

1 solution

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