66Is it so obvious. Probably, one should remember that the Intermediate value property holds for closed intervals.
This is crackable but nice:
f:(a,b) \rightarrow \mathbb{R} is continuous. Prove that given x_1, x_2, x_3,...,x_n in (a,b), there exists x_0 \in (a,b) such
that f(x_0) =\frac{1}{n} \left(f(x_1)+f(x_2)+f(x_3)+...+f(x_n) \right)
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7 Answers
1
f(x0) = 1/n[f(x1)+f(x2)+.................f(xn)]
is the arithematic mean of [f(x1)+f(x2)+.................f(xn)]
And an Arithematic Mean always lies in between the maximum and minimum value
ie. x0 belongs to (a,b)
1THAT'S WHY I SAID IT IS OBVIOUS.
IS THERE SOMETHING ELSE IN QUESTION? WHICH I AM NOT UNDERSTANDING.
62I repeat a famous quote that a prof once told
"What is obvious to you might not be obvious to me"
66
341In other words the IMV is being applied to the interval ________ ?
