Using Newton's Method, what is the formula for determining the nth term in a sequence of approximations to a solution of a polyn
1 answer:
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Answer:
Option C.
Step-by-step explanation:
Note: In the given function the power of x should be 2 instead of 4, otherwise all options are incorrect.
Consider the given function is
If , then
is an even function.
If , then
is an odd function.
Now, substitute x=-x in the given function.
So, the given function not an odd function. It means it is an even function.
To check whether the given function is odd, we have to determine whether is equivalent to
.
Therefore, the correct option is C.
Answer:
Step-by-step explanation:
Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.
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Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².
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Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).
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Taken altogether, these factors give us the function ...
