If f(2)=4 and f'(x)>2 for 2<x<11, how small can f(11) be?
Thanks!
Note that f satisfies the conditions for the mean value theorem to be
applicable on [2,11].
Suppose f(11)<22, then by the mean value theorem there exists a c in (2,11)
such that:
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.
RonL
Hello, asnxbbyx113!
Did you make a sketch?
If $\displaystyle f(2)=4$ and $\displaystyle f'(x) \geq 2$ for $\displaystyle 2 \leq x \leq 11$, how small can $\displaystyle f(11)$ be?
We consider the graph of $\displaystyle f(x)$ from $\displaystyle x = 2$ to $\displaystyle x = 11$.
SInce $\displaystyle f(2) = 4$, the graph starts at $\displaystyle (2,4)$.Code:| (2,4) | * | | - + - + - - - - - - - | 2
From $\displaystyle x = 2$ to $\displaystyle x = 11$, the slope is $\displaystyle \text{at least 2.}$
The very "lowest" graph would be straight line from $\displaystyle (2,4)$ with slope $\displaystyle 2$.
. . Its equation is: .$\displaystyle f(x) \:=\:2x$
It would look like this:Code:| | * (11,22) | * : | * : | * : | * : | * : | * : | * : | : : | : : - | - + - - - - - - + - - | 2 11
Hence, the least value for $\displaystyle f(11)$ is: .$\displaystyle f(11) = 22$
Well I've always known about the use of the MVT, but I thought it
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
RonL