If f(2)=4 and f'(x)>2 for 2<x<11, how small can f(11) be?
applicable on [2,11].
Suppose f(11)<22, then by the mean value theorem there exists a c in (2,11)
f'(c)=(f(11)-f(2))/9 = (f(11)-4)/9,
but if f(11)<22 this is <2, which contradicts the assumption that f'(x)>=2.
Now let f(x)=4+2(x-2), then f'(x)=2 for x in (2,11), and f(11)=22.
Therefore f(11) can be as small as 22 and no smaller.
Did you make a sketch?
If and for , how small can be?
We consider the graph of from to .
SInce , the graph starts at .Code:| (2,4) | * | | - + - + - - - - - - - | 2
From to , the slope is
The very "lowest" graph would be straight line from with slope .
. . Its equation is: .
It would look like this:Code:| | * (11,22) | * : | * : | * : | * : | * : | * : | * : | : : | : : - | - + - - - - - - + - - | 2 11
Hence, the least value for is: .
inappropriate for the level of knowlege of the student involved, as
it probably is here also, which is why Soroban is thought a better helper
than both of us .
A MVT proof is more appropriate to a first Real Analysis course than to
any course calling itself Calculus X.
It seems some of your tendency to answer questions with methods
above the heads of the asker has rubbed of on me