Suppose a function f: [0, 10] -> R is continuous and differentiable everywhere in its domain. If f(10) = 19, and |f'(x) - 5| ≤ 4 for all x in the domain, what is the maximum value of f(0)?
from
|f'(x)-5|<=4
-4<=f'(x)-5<=4
1<=f'(x)<=9
thus f'(x)>0 and is strictly increasing on [0,10]
we are given f(10)=19, let f(0)=a, then from the mean value theorem we have that there exists a value x on [0,10] for which f'(x)=[f(10)-f(0)]/(10-0)=(19-a)/10 but from above we know this must satisfy
1<=f'(x)<=9
Thus
1<=(19-a)/10<=9
10<=19-a<=90
-171<=a<=71
thus the maximum possible value for f(0) is 9
For an explicit example take f(x)=9+x
f(0)=9, f(10)=19 and f'(x)=1 which satisfies 1<=f'(x)<=9
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Posted by Daniel
on 2019-07-15 12:44:04 |
solution