Suppose that we know that f(x) is a continuous and differentiable function on [2,14]. Also, suppose we know that f(2)=9 and f'(x) is greater than or equal to -12. What is the smallest possible value of f(14)?

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Suppose that we know that f(x) is a continuous and differentiable function on [2,14]. Also, suppose we know that f(2)=9 and f'(x) is greater than or equal to -12. What is the smallest possible value of f(14)?

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would this be a straight line from 2,14 running at a slope of 12?
(I'm supposed to use the mean value theorem, or at least that's what it says on the worksheet.)
(I'm supposed to use the mean value theorem, or at least that's what it says on the worksheet.)

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@amistre64 - I have no idea! I'm complete lost on this one. Don't even know where to start.
f'(2) =>12....6x? maybe
Well, if f'(x) = -12 throughout the interval then it would be a straight line from (2,9) with a slope of -12. So that would be the minimum possible value given what we know.
Sorry, @polpak - how do we know that?
ahh,....(-)12 ... that was hiding in the corner
We are given that f'(x) is greater than or equal to -12.
So if it maintained the smallest possible derivative over the whole interval we would have a line with slope of -12.
i got the smallest possible value is -135 for f(14)
thats what I thought :)
but with a 12 lol
9-12*12 = 9-144 = -135
yay thats what i got except differently
-12<=[f(2)-f(14)]/[2-14]
Does that make sense Chap?
How do you figure what f(14) is?
chap you can also solve the above for f(14) resulting in f(14)>=(-135)
I used the mean value thm just like you said
Right. But I have -12<= 9 - f(14)
Oops wait
I have -12 <= 9 - f(14) ----------- -12
multipliy -12 on both sides don't forget to flip the inequality when you multipliy or divide by a negative
Sorry )= I don't understand how to do that.
\[-12 \le \frac{9- f(14)}{-12} \implies 144 \ge 9-f(14)\]
how do you solve x/5=4
\[-4a < b \implies a \gt \frac{b}{-4}\]
Okay. But how does 135 >= - f(14) solve for f(14)?
Multiply by -1
now multipliy both sides by negative 1 don't forget to flip
135 <= f(14)
right!
wait where's the negative?
That's the answer?
-135 <= f(14)
ok so we have f(14)>=-135 this means the smallest that f(14) can be is -135
OH! I thought f(14) would be positive. Thank you so much!
It might be positive!
We only know that the smallest it can be is -135. It could be very very large.
it could be 1111111111111111111111
over 9000!
5 trillion billion if that means anything

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