At which values of x does the function F(x) have a vertical asymptote? Check all that apply. F(x)=2/3x(x-1)(x 5).
1 answer:
The function F(x) have a vertical asymptote at x =0 and x =1.
Given that
The given Function;
We have to determine
At which value the function F(x) has a vertical asymptote.
According to the question
The given Function;
The function is undefined when the denominator becomes 0. Find values of x for which the denominator is 0:
Then,
The value of x is 0 and 1.
Hence, the function F(x) have a vertical asymptote at x =0 and x =1.
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Let
where we assume |r| < 1. Multiplying on both sides by r gives
and subtracting this from gives
As n → ∞, the exponential term will converge to 0, and the partial sums will converge to
Now, we're given
We must have |r| < 1 since both sums converge, so
Solving for r by substitution, we have
Recalling the difference of squares identity, we have
We've already confirmed r ≠ 1, so we can simplify this to
It follows that
and so the sum we want is
which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?