Let f(x) and g(x) are two differentiable function on R satisfying 3 1 +1-x g(t)dt and g(x)= =X- -}ƒ(t)at. f(t)dt. If minimum distance between f(x) and g(x) is d, then 0 f(x) √2 dis = 2 X 0

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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I found derivatives of f(x) using Newton- Leibniz and then g'(x) too. I equated both of them, put g(x) back into f'(x) because the expression of f'(x) had a term of g(x). Then, I differentiated the final expression again in hopes of getting the actual function of g(x). However, doing all of this got me nowhere. 

Please provide a solution with explanation of the thought process too.

Let f(x) and g(x) are two differentiable function on R satisfying
3
f(x)=+1-xg(t)dt
2
√√2d is
X
-xſg(t)dt
1
and g(x)= x − [ f(t)dt . If minimum distance between f(x) and g(x) is d, then
0
Transcribed Image Text:Let f(x) and g(x) are two differentiable function on R satisfying 3 f(x)=+1-xg(t)dt 2 √√2d is X -xſg(t)dt 1 and g(x)= x − [ f(t)dt . If minimum distance between f(x) and g(x) is d, then 0
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