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)]/(100)=(19a)/10 but from above we know this must satisfy
1<=f'(x)<=9
Thus
1<=(19a)/10<=9
10<=19a<=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

Posted by Daniel
on 20190715 12:44:04 