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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.)

@amistre64 - I have no idea! I'm complete lost on this one. Don't even know where to start.

f'(2) =>12....6x? maybe

ahh,....(-)12 ... that was hiding in the corner

We are given that f'(x) is greater than or equal to -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

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