Question

a friend runs up to you and excitedly explains that she has found a function g with the following properties:
g is continuous on [0,1] and differentiable on (0,1)
g(0)=1 and g(1)=5
g'(x) is < or equal to 3 for all x in (0,1)

explain why you doubt your friends claim

Answer 1

@thomaster can you please help?

Answer 2

Suppose \(g'(x)\) takes on the highest value it can for the interval, so that \(g'(x)=3\).

Integrating, you have
\[g(x)=\int g'(x)~dx=3x+C\]
You know that \(g(0)=1\), so
\[g(0)=1=3(0)+C~~\Rightarrow~~C=1\]
But then
\[g(1)=3(1)+1=4\not=5\]

Answer 3

Actually, 4 < 5.

Answer 4

Both \(4<5\) and \(4\not=5\) tell us that 4 isn't equal to 5

Answer 5

Sorry, I thought 4 < 5 was what you wanted to say.

Answer 6

I certainly could have, but the point is that \(g(1)=5\) can't be satisfied even if the function assumes maximum rate of change.

Answer 7

Why can't it? (Without using the fact that 4 < 5, that is.)

Answer 8

Imagine that in your example \(g(1) = 6 \neq 5\). Would that imply that \(g(1) = 5\) cannot be satisfied?

Answer 9

basically the friend is wrong because she claims that g(1)=5 but g(1) actually equals 4 not 5?

Answer 10

Regarding your example, if \(g(1)=6\) then certainly \(g(1)\not=5\); otherwise \(g\) is not a function.

Answer 11

\(\huge\ mean\ value\ theorem\)

Answer 12

Here are the cases:

\(g'(x)\le3\) for all \(x\) in the interval. If \(g'\) is negative then of course the function is decreasing, so any value of \(g(x)\) will be smaller than the initial value \(g(0)=1\), so we can ignore those.

If \(g'=0\), there's no change in \(g\), so \(g(0)=g(1)=1\).

If \(g'>0\), we can assume the highest rate of change \(g'=3\) throughout the interval, then you get the result I wrote earlier.

Now that I think about it though, is it possible that \(g\) is not a linear function, and that it could be a piecewise function that's differentiable and still satisfies \(g(1)=5\)?

Answer 13

Thanks mathmate, I never saw a good example of the mean value theorem.

Answer 14

i'm still confused but i doubt g is a piecewise function

Answer 15

@tiffmahoney To be very clear, by the mean value theorem, because g(0)=1 and g(1)=5 there exists a point on (0, 1) in which the derivative is (5-1)/(1-0) = 4, but that's not possible.

Answer 16

That's not possible because 4 is not < or equal to 3.

Answer 17

okay, so you plug the corresponding numbers into a & b of the formula f(b)-f(a)/b-a and if that result doesnt satisfied g'(x) then is incorrect?

Answer 18

g(x) is continuous and differentiable, so piecewise and step functions are excluded.
So we recall that Mean Value Theorem applies if and only if the function is continuous and differentiable within a closed interval.

Answer 19

Yes.

Answer 20

thanks for the help guyss!! i have one more problem i really need help with too, it would be great if one of you can help

Answer 21

I'll try.

Answer 22

@mathmate we can't be quick to exclude all piecewise functions. For instance,
\[h(x)=\begin{cases}x^2&\text{for }x\ge0\\0&\text{for }x<0\end{cases}\]
is still continuous and differentiable everywhere.

I was wondering if there's any function \(g\) that could follow this sort of structure...

Answer 23

@SithsAndGiggles
Agree! I over-generalized. I should have said to exclude step functions and most piecewise functions unless they are shown to be continuous and differentiable everywhere within the closed interval.
Yes, it is possible, but by virtue of the mean value theorem, there still has to be a point somewhere within the interval having a slope of 4. The statement that g(x) has a MAXIMUM slope of 3 within the interval clearly conflicts with the conclusions of the mean value theorem.