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What is the discontinuity and zero of the function f(x) = 3x^2 + x - 4 / x-1

Question 3 options:
1) Discontinuity at (1, 7), zero at ( negtive four thirds , 0)
2) Discontinuity at (−1, 1), zero at ( negative four thirds , 0)
3) Discontinuity at (1, 7), zero at ( four thirds , 0)
4) Discontinuity at (−1, 1), zero at ( four thirds , 0)

User Lepton
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1 Answer

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Ans: Option (1) Discontinuity at (1, 7), zero at ( negative four thirds , 0)

Step-by-step explanation:
Given function:

f(x) = (3x^2 + x - 4)/(x-1)

Now if we plug in the x = 1, we would have discontinuity as function goes to infinity.

Now for f(1):

f(x) = (3x^2 + x - 4)/(x-1) \\ f(x) = (3x^2 + 4x - 3x - 4)/(x-1) \\ f(x) = (x(3x+4)-1(3x+4))/(x-1) \\ f(x) = ((x-1)(3x+4))/((x-1)) \\ f(x) = 3x+4 \\ now ~ insert ~ x=1: \\ f(1) = 3(1) + 4 = 7 \\ It~means~discontinuity~at~(1,7).\\ Now~let~us~find~zeros.\\ put~ f(x) = 3x+4 ~ equals~ to ~zero. \\ =\ \textgreater \ ~3x+4 = 0 \\ =\ \textgreater \ ~ x = (-4)/(3) \\ Hence~zeros=( (-4)/(3), 0)

User Zmarties
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