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Property 7

consider a re...

Question

consider a real valued function such that f(x+2)=-f(2-x) then prove that the value of integration of f(x) from -3 to 7 is equal to 0.

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Solution

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f (x) + 2 f (\frac{932}{x}) = 3 x

\neq

f (x)

∣ ∣ f (x) + x^{2} + 1 ∣ ∣ \geq | f (x) | + ∣ ∣ x^{2} + 1 ∣ ∣

f (x) \leq 0

x

5 \sum r = 1 (1 + f (r))

f (x)

x

f (x) = \sqrt{x + 2} + \sqrt{7 - x}

f (x)

f^{'} (x)

f^{''} (x)

x

f (0) = 0, f (1) = 2, f (2) = 1

f (3) = - 3

f (x) \geq 1 \forall x \in R

f (2)

1

f (3) = 3

f (5)

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