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Sum of squares of average change is less than

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB,

I have no good ideas on how to go about solving the following:

Let $f:[a,b]\to\mathbb R$ and $g:[a,b]\to\mathbb R$ be real values functions both of which are differentiable in $(a,b)$. Show that there is an $x\in(a,b)$ such that $$\left(\frac{f(b)-f(a)}{b-a}\right)^2+\left(\frac{g(b)-g(a)}{b-a}\right)^2\leq (f'(x))^2+(g'(x))^2$$

Please help.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: Sum of squares of average change is less than ...

Hello MHB,

I have no good ideas on how to go about solving the following:

Let $f:[a,b]\to\mathbb R$ and $g:[a,b]\to\mathbb R$ be real values functions both of which are differentiable in $(a,b)$. Show that there is an $x\in(a,b)$ such that $$\left(\frac{f(b)-f(a)}{b-a}\right)^2+\left(\frac{g(b)-g(a)}{b-a}\right)^2\leq (f'(x))^2+(g'(x))^2$$

Please help.
Here's an idea to get you started. Define a path in 2-dimensional space by the function $h:[a,b] \to \mathbb{R}^2$ given by $h(t) = \bigl(f(t),g(t)\bigr)\ (a\leqslant t\leqslant b).$ Then the distance between the endpoints is less than (or equal to) the length of the path.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: Sum of squares of average change is less than ...

Here's an idea to get you started. Define a path in 2-dimensional space by the function $h:[a,b] \to \mathbb{R}^2$ given by $h(t) = \bigl(f(t),g(t)\bigr)\ (a\leqslant t\leqslant b).$ Then the distance between the endpoints is less than (or equal to) the length of the path.
Here's an idea to get you started. Define a path in 2-dimensional space by the function $h:[a,b] \to \mathbb{R}^2$ given by $h(t) = \bigl(f(t),g(t)\bigr)\ (a\leqslant t\leqslant b).$ Then the distance between the endpoints is less than (or equal to) the length of the path.
So the above says that:

$\sqrt{(f(b)-f(a))^2+(g(b)-g(a))^2}\leq \displaystyle\int_a^b\left[\sqrt{(f'(x))^2+(g'(x))^2}\right]dx$. Now we use the fact that if $h:[a,b]\to\mathbb R$ is a continuous function then $\int_a^bh(x)dx\leq h(x_0)(b-a)$ for some $x_0\in (a,b)$.

Is this correct?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: Sum of squares of average change is less than ...

So the above says that:

$\sqrt{(f(b)-f(a))^2+(g(b)-g(a))^2}\leq \displaystyle\int_a^b\left[\sqrt{(f'(x))^2+(g'(x))^2}\right]dx$. Now we use the fact that if $h:[a,b]\to\mathbb R$ is a continuous function then $\int_a^bh(x)dx\leq h(x_0)(b-a)$ for some $x_0\in (a,b)$.

Is this correct?
That's exactly what I was thinking of. (Nod)