1IS IT NOT OBVIOUS?
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"
66Is it so obvious. Probably, one should remember that the Intermediate value property holds for closed intervals.
341In other words the IMV is being applied to the interval ________ ?
