which of the following is an odd function? a. f(x)= xcosx b.f(x)= xsinx c.f(x)= e ^cosx d.f(x)=sin^2x
Test and see. What's the definition of an odd function?
my answer again is letter c i dont know y.
if a or b my answer the wiki is correct
By definition, a function f is odd if \[ f(-x) = -f(x) \] The function in c is not odd ...
you need to check to see if \[f(-x)=f(x)\] for even or if \[f(-x)=-f(x)\] for odd and there is no shame in trying it with numbers if the x's get confusing
it won't be a proof, but it will give you an idea of what is going on. then you can do it with a variable
...because \[ \cos(-x) = \cos(x) \] hence \[ e^{\cos(-x) } = e^{\cos(x)} \]
odd x odd = even even x odd= odd even x even= even
yun!
ic all your answer is correct but a or b are the same right? i only need to choose 1 letter and it is letter D
No. The function is D is not odd, because \[ \sin^2(-x) = (\sin(-x))^2 = (-\sin x)^2 = \sin^2 x \]
hence the function in D and C are both even functions: \[ f(-x) = f(x) \]
how could i know a and b are odd? i feel sinx and cosx are the same
OMG its LETTER B because cos x is an even function
No. Try and work it out by explicitly evaluating expressions.
Geometrically, the graph of an odd function has rotational symmetry with respect to the origin, meaning that its graph remains unchanged after rotation of 180 degrees about the origin. Examples of odd functions are x, x3, sin(x), sinh(x), and erf(x).
Yes. But stop avoiding doing the calculation. Take each of the functions in turn and evaluate f(-x). See if it equal to -f(x) or not. If it is, then f(x) is odd. That is how you answer this question.
f(x) = -sinx is odd function so f(x) = xcosx
f(x) = x.cos(x) is an odd function yes. Why? Because f(-x) = (-x).cos(-x) = -x.cos(x) , because cos(-x) = cos(x) = -f(x)
its like -f(x) = f(-x) thank your mister james
By definition, a function f(x) is odd if f(-x) = -f(x)
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