Which statement is true about the discontinuities of the function f(x)?
1 answer:
Answer:
(a) There are asymptotes at x=3/2 and x=-1/3
Step-by-step explanation:
The denominator zeros can be found by factoring:
f(x) = (x +1)/((2x -3)(3x +1))
Neither of the denominator factors is cancelled by the numerator factor, so each represents a vertical asyptote, not a function hole.
The asymptotes are at the values of x where the denominator is zero:
2x -3 = 0 ⇒ x = 3/2
3x +1 = 0 ⇒ x = -1/3
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option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).
<u>Step-by-step explanation:</u>
We have , The given expression as Three-fourths x = negative 6 , which can be written as . Now in order to solve this equation in one step , we must notice that coefficient of x must be 1 but it's
, Let's make coefficient of x as 1 by multiplying both side of equations by 4/3:
⇒
⇒
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Therefore, x= -8 & correct option to solve the equation Three-fourths x = negative 6 for x in one step is <u>option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).</u>