What Is the Least Possible Value for the Function f(4)?

[madah12]

Homework Statement

f(1)=10
f'(x) >= 2 for every x in R
1<x<4
find the least possible value for f(4)

Homework Equations

The Attempt at a Solution

f is differentiable on R therefore continuous there exist a C such that
f(4)-10 / 3 = f'(C) >=2
f(4)-10 / 3 >= 2
f(4) >= 16 , so the least value is 16
I want to know if this second approach is correct because my teacher said it is correct but you must present a theorem to write it
I want to say that suppose we make g (x) = 2x + 10 so since the least value for f(4) is if f' is always the least possible value which is 2 and for f' to always equal two it must be a straight line but I only know that is a line from x=1 so i must make 10 my y intercept 4 is actually 3 units away from 1 so my line starts from 0 which equals 2 in the function so 4 is 3 so f(3) = 6+10 = 16
how can I make more reasonable mathematics out of this?

 

[micromass]

So basically, you want to show that f(x)\geq 2x+10.

For x=1, this is true.
Assume that there is a x>1 such that f(x)<2x+10. Now apply the mean value theorem to 1 and x...

 

[madah12]

So basically, you want to show that f(x)\geq 2x+10.

For x=1, this is true.
Assume that there is a x>1 such that f(x)<2x+10. Now apply the mean value theorem to 1 and x...

I don't think so f(1) =10 but 2*1 +10 = 12 12>10 but that isn't what I am saying because I am using 2x+10 only starting from 1 I am using another axis such that when x=0 in the axis of the original function x=0 in 2x+10.

 

[Mark44]

The smallest value of f' is 2, which is what f'(x) >= 2 means. This also means that the graph of f has to lie on or above the line segment whose slope is 2, that passes through (1, 10). What is the y-coordinate on this line when x = 4? This will be the smallest possible value for f(4).

 

[madah12]

y=2x+b
2+b=10
b=8
y=2x+8
=2*4+8=16
but is there a theorem or a definition that tells me that in order to get the least value I need to use a line with slope two? I mean I get it with common sense but there is no theorem that says if f' is always two f will be the least value which is my problem.

 

[Mark44]

You could probably do something with the intermediate value theorem. The basic idea is that the graph of f has to lie on or above the line between (1, 10) and (4, 16). Otherwise, if the graph of f were below this line, the IVT guarantees that for some c such 1 <= c <= 4, f'(c) = 2. This means that for some x > c, f'(c) > 2, which you can't have.