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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)

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

Explanation:
Given function:
[tex]f(x) = \frac{3x^2 + x - 4}{x-1} [/tex]

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

Now for f(1):
[tex]f(x) = \frac{3x^2 + x - 4}{x-1} \\ f(x) = \frac{3x^2 + 4x - 3x - 4}{x-1} \\ f(x) = \frac{x(3x+4)-1(3x+4)}{x-1} \\ f(x) = \frac{(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 = \frac{-4}{3} \\ Hence~zeros=( \frac{-4}{3}, 0)[/tex]

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