Suppose that f(0) = -2 and f'(x) \le 3 for all values of x. Use the Mean Value Theorem to determine how large f(4) can possibly be. Answer: f(4) \le
If f(x) is continuous on [0,4], and differentiable on (0,4), Then there exists a tangent line in that interval, with the same slope as the secant line connecting the points at x=0 and x=4.
\[\large\rm \frac{f(4)-f(0)}{4-0}=f'(x),\qquad\qquad x\in(0,4)\]
\[\large\rm \frac{f(4)-(-2)}{4}=f'(x)\le 3\]So it follows directly that,\[\large\rm \frac{f(4)+2}{4}\le3\]And then solve for f(4) from there :) Any confusion on those steps?
The final answer would be 18? I'm confuse a little bit.
18? Hmm that doesn't sound right... Do you understand how the Mean Value Theorem works? If you take an interval of a function, which has no holes or funny business going on in that interval, Then the slope of the line connecting the ends points of the interval, must be equal to the slope of the derivative somewhere in that interval. I don't think I can draw a picture to illustrate it, on laptop right now :p hmm
In fact, I don't know about MVT and its my homework question I could not go to math center today and I have to do my homework after 2h I don't know what to do. Thanks and I appreciate ur help.
So you have some line which passes through f(0) and f(4). Maybe we call it y. Since it's a straight line, it has the form y=mx+b. We're only concerned with this m, the slope of the line. Remember your slope formula? \[\large\rm m=\frac{y_2-y_1}{x_2-x_1}\]But in calculus we're now writing it using function notation,\[\large\rm m=\frac{f(x_2)-f(x_1)}{x_2-x_1}\] We call this a secant line, a line passing through 2 or more points. Our Mean Value Theorem tells us that there is a tangent line in this interval with the same slope as that secant line. Lemme try to draw an example XD
|dw:1448324219912:dw|Oh boy -_-
It doesn't matter what the curve looks like in between those two points, the Mean Value Theorem guarantees a tangent line will exist with the same slope as that secant line.|dw:1448324308508:dw|
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