Question 1209870
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Problem 1


If f(x) is an even function, then f(-x) = f(x) for all x in the domain.
This produces symmetry about the y axis.

Since functions g and h are even, we know that
g(-x) = g(x) and h(-x) = h(x) 


Then,
k(x) = f(x)+g(x)+h(x)
k(-x) = f(-x)+g(-x)+h(-x) ....... replace each x with -x
k(-x) = f(x)+g(x)+h(x) ........ apply definitions shown above
k(-x) = k(x)
Therefore k(x) is <font color=red>even</font>


You can use a graphing tool like <a href="https://www.desmos.com/calculator">Desmos</a> or <a href="https://www.geogebra.org/calculator">GeoGebra</a> to help confirm the answer.
For instance you can form these functions
f(x) = x^2
g(x) = 10x^4-7
h(x) = 8x^6+3x^2
A polynomial is an even function when all exponents are even numbers. 
Think of the -7 as -7x^0 in the g(x) function.


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Problem 2


If f(x) is an odd function then f(-x) = -f(x) for all x in the domain.
g(-x) = -g(x) is the case here as well since we're told g(x) is also odd.
h(-x) = h(x) since h(x) is even.


k(x) = f(x)*g(x) + h(x)
k(-x) = f(-x)*g(-x) + h(-x)
k(-x) = -1*f(x)*(-1)*g(x) + h(x)
k(-x) = (-1)*(-1)*f(x)*g(x) + h(x)
k(-x) = f(x)*g(x) + h(x)
k(-x) = k(x)
This proves that k(x) is <font color=red>even</font>


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Problem 3


f = even function
g,h = odd functions


f(-x) = f(x)
g(-x) = -g(x)
h(-x) = -h(x)


k(x) = f(x)*( g(x)+h(x) )
k(-x) = f(-x)*( g(-x)+h(-x) )
k(-x) = f(x)*( -g(x)-h(x) )
k(-x) = f(x)*(-1)*( g(x)+h(x) )
k(-x) = -f(x)*( g(x)+h(x) )
k(-x) = -k(x)
We have shown that k(x) is <font color=red>odd</font>


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Problem 4


f,h = odd functions
g = even


f(-x) = -f(x) and h(-x) = -h(x)
g(-x) = g(x)


k(x) = f(x)*g(x)*h(x)
k(-x) = f(-x)*g(-x)*h(-x)
k(-x) = -f(x)*g(x)*(-1)*h(x)
k(-x) = (-1)*(-1)*f(x)*g(x)*h(x)
k(-x) = f(x)*g(x)*h(x)
k(-x) = k(x)
k(x) is an <font color=red>even</font> function.


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Answers:
1. <font color=red>even</font>
2. <font color=red>even</font>
3. <font color=red>odd</font>
4. <font color=red>even</font>
In other words, answer 3 is the only "odd" while the others are "even".
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