An even function can be reflected about the y axis and map onto itself
example: y=x^2
an odd function can be reflected about the origin and map onto itself
example: y=x^3
a simple test is the following
if f(x) is even then f(-x)=f(x)
if f(x) is odd then f(-x)=-f(x)
so
even function
subsitute -x for each and see if we get the same function
remember to fully expand these
g(x)=(x-1)^2+1=x^2-2x+1+1=x^2-2x+2 is the original one
g(x)=(x-1)^2+1
g(-x)=(-x-1)^2+1
g(-x)=(1)(x+1)^2+1
g(-x)=x^2+2x+1+1
g(-x)=x^2+2x+2
not same because the original has -2x
not even
g(x)=2x^2+1
g(-x)=2(-x)^2+1
g(-x)=2x^2+1
same, it's even
g(x)=4x+2
g(-x)=4(-x)+2
g(-x)=-4x+2
not the same, not even
g(x)=2x
g(-x)=2(-x)
g(-x)=-2x
not same, not even
g(x)=2x²+1 is the even function
Answer: NONE
<u>Step-by-step explanation:</u>
Consider that m is the degree of the numerator and n is the degree of the denominator.
The rules for horizontal asymptote (H.A.) are as follows:
If m > n then no H.A. (use long division to find the slant asymptote)
If m = n then H.A. is y = leading coefficient of numerator/leading coefficient of denominator
If m < n then H.A. is y = 0
Given: g(x) = 5x⁵/(x³ - 2x + 1)
--> m = 5, n = 3
Since m > n then there is no H.A.
Answer:
f(g(-64)) = -190
Step-by-step explanation:
The functions are not well written.
Let us assume;
f(x) = x+1
g(x) = 3x+1
f(g(x)) = f(3x+1)
Replace x with 3x+1 in f(x)
f(g(x)) = (3x+1) + 1
f(g(x)) = 3x + 2
f(g(-64)) = 3(-64) + 2
f(g(-64)) = -192+2
f(g(-64)) = -190
<em>Note that the functions are assumed but same method can be employed when calculating composite functions</em>
