If f and g are differentiable functions in [0,1] satisfying f(0)=2=g(1), g(0)=0 and f(1)=6, then for some c∈[0,1]:
Solutions · 1
Official solution
Since, f and g both are continuous function on \[\left[ {0,1} \right]\]
and differentiable on \[\left( {0,1} \right)\] then \[\exists c \in \left( {0,1} \right)\] such that
\[f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4\]
and \[g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2\]
Thus, we get \[f'\left( c \right) = 2g'\left( c \right)\]