Let the function f (x) $$=$$ $$3$$ $$sin$$ x – $$4$$ $$cos$$ x $$+$$ log $$($$|x| $$+$$ $$1+x2$$ be defined on the interval [$$0$$, $$1$$]. The odd extension of f (x) to the interval [– $$1$$, $$1$$] is

Question

Let the function f (x) $$=$$ $$3$$ $$sin$$ x – $$4$$ $$cos$$ x $$+$$ log $$($$|x| $$+$$ $$1+x2$$ be defined on the interval [$$0$$, $$1$$]. The odd extension of f (x) to the interval [– $$1$$, $$1$$] is

Infinity Learn · Accepted Answer

To make f (x) an odd function in the interval [–$$1$$, $$1$$], we $$re-define$$ f (x) as follows :$$f(x)=f(x)$$,    $$0$$≤x≤$$1$$−$$f($$−$$x)$$,    −$$1$$≤x<$$0=3sin$$⁡x−$$4cos$$⁡$$x+log$$⁡|x|$$+1+x2$$,    $$0$$≤x≤$$1$$−−$$3sin$$⁡x−$$4cos$$⁡$$x+log$$⁡|x|$$+1+x2$$,    −$$1$$≤x<$$0=3sin$$⁡x−$$4cos$$⁡$$x+log$$⁡|x|$$+1+x2$$,$$0$$≤x≤$$13sin$$⁡$$x+4cos$$⁡x−log⁡|x|$$+1+x2$$,−$$1$$≤x<$$0Thus$$, the odd extension of f (x) to the interval [–$$1$$, $$1$$] $$is3sin$$⁡$$x+4cos$$⁡x−log⁡|x|$$+1+x2$$

Let the function f (x) = 3 sin x – 4 cos x + log (|x| + $\sqrt{1 + x^{2}})$ be defined on the interval [0, 1]. The odd extension of f (x) to the interval [– 1, 1] is

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Let the function f (x) = 3 sin x – 4 cos x + log (|x| + 1+x2 be defined on the interval [0, 1]. The odd extension of f (x) to the interval [– 1, 1] is

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Let the function f (x) = 3 sin x – 4 cos x + log (|x| + $\sqrt{1 + x^{2}})$ be defined on the interval [0, 1]. The odd extension of f (x) to the interval [– 1, 1] is