Notice that f ( x ) {\displaystyle \displaystyle f(x)} is differentiable on [ 0 , 4 ] {\displaystyle \displaystyle [0,4]} and hence it is continuous there. Thus, by the mean value theorem, there exists a point c ∈ [ 0 , 4 ] {\displaystyle \displaystyle c\in [0,4]} such that
f ′ ( c ) = f ( 4 ) − f ( 0 ) 4 − 0 {\displaystyle \displaystyle f'(c)={\frac {f(4)-f(0)}{4-0}}} .
We know that on [ 0 , 4 ] {\displaystyle \displaystyle [0,4]} that f ′ ( x ) ≥ 3 {\displaystyle \displaystyle f'(x)\geq 3} . Since c ∈ [ 0 , 4 ] {\displaystyle \displaystyle c\in [0,4]} and f ( 0 ) = 10 {\displaystyle \displaystyle f(0)=10} , we have that
3 ≤ f ′ ( c ) = f ( 4 ) − f ( 0 ) 4 − 0 = f ( 4 ) − 10 4 {\displaystyle \displaystyle 3\leq f'(c)={\frac {f(4)-f(0)}{4-0}}={\frac {f(4)-10}{4}}} .
Solving gives 22 ≤ f ( 4 ) {\displaystyle \displaystyle 22\leq f(4)} and hence the minimum f ( 4 ) {\displaystyle \displaystyle f(4)} could be is 22.