SOLUTION: Hi all, I dont quite understand what this questions is asking: Determine whether the following functions y = xsin3x, {{{y = x*sin(3x)}}}, {{{y = x^2 + e^x}}}, {{{y = x^3*cosx}}}

Algebra ->  Vectors -> SOLUTION: Hi all, I dont quite understand what this questions is asking: Determine whether the following functions y = xsin3x, {{{y = x*sin(3x)}}}, {{{y = x^2 + e^x}}}, {{{y = x^3*cosx}}}       Log On


   



Question 205201: Hi all, I dont quite understand what this questions is asking:
Determine whether the following functions y = xsin3x, y+=+x%2Asin%283x%29, y+=+x%5E2+%2B+e%5Ex, y+=+x%5E3%2Acosx are even, odd or none of these.
Can anyone help me with solving it and explain how to do it please?
Thanks, -Nick.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+x%2Asin%283x%29, y+=+x%5E2+%2B+e%5Ex, y+=+x%5E3%2Acosx

The graphs of even functions are symmetrical with respect to the
y-axis.  Odd functions are symmetrical with respect to the origin.

There are two ways to show even, odd, or neither.  The first way is
graphical. The second way is algebraic.  I'll do the graphical way
first:

The graph of y+=+x%2Asin%283x%29

graph%28400%2C400%2C-10%2C10%2C-10%2C10%2Cx%2Asin%283x%29%29

It is symmetrical with respect to the y-axis, so it is an even function.

The graph of y+=+x%5E2+%2B+e%5Ex

graph%28400%2C400%2C-10%2C10%2C-10%2C10%2Cx%5E2+%2B+e%5Ex%29

It is not symmetrical with respect to the y-axis or with respect to the
origin so it is neither even nor odd.

The graph of y+=+x%5E3%2Acos%28x%29

graph%28400%2C400%2C-10%2C10%2C-300%2C300%2Cx%5E3%2Acos%28x%29%29

It is symmetrical with respect to the origin, so it is odd.

------------

How to tell algebraically whether the graph will be symmetrical
with the y-axis (even), symmetrical with the origin (odd), or
not symmetrical with respect to either the y-axis or the origin
(neither):

Rules for the algebraic method:

Substitute -x for x and simplify.  If you get the same right side
you started with after simplifying, the function is even.

If you get the original right side multiplied by -1, then the 
function is odd.

If you get neither, the function is neither even nor odd.

----------------------

y+=+x%2Asin%283x%29

replace the x's by (-x)

y+=+%28-x%29%2Asin%283%28-x%29%29

y+=+-x%2Asin%28-3x%29

Use the fact that sin%28-phi%29=-sin%28phi%29

y+=+-x%2A%28-sin%283x%29%29

Two negatives make a positive

y+=+x%2Asin%283x%29

This is the same as the original right side so the
function is even.

----------------

 y+=+x%5E2+%2B+e%5Ex, 

Replace the x's by (-x)

y+=+%28-x%29%5E2+%2B+e%5E%28-x%29

y+=+x%5E2+%2B+e%5E%28-x%29

That is neither the same as the original right side,
nor is it the original multiplied by -1,
so it is neither an even nor an odd function.

y+=+x%5E3%2Acos%28x%29

y+=+%28-x%29%5E3%2Acos%28-x%29

Use the fact that cos%28-phi%29=cos%28phi%29

y+=+-x%5E3%2Acos%28x%29

That is the original right side multiplied by -1, so
the function is odd.

Edwin