To determine the discontinuity and zero of the function [tex]\( f(x) = \frac{3x^2 + x - 4}{x - 1} \)[/tex], we need to perform the following steps:
1. Find the Discontinuity:
The discontinuity of the function occurs when the denominator is equal to zero, as this would make the function undefined. For the function [tex]\( f(x) = \frac{3x^2 + x - 4}{x - 1} \)[/tex], let's set the denominator [tex]\(x - 1\)[/tex] equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x - 1 = 0
\][/tex]
[tex]\[
x = 1
\][/tex]
Thus, there is a discontinuity at [tex]\(x = 1\)[/tex]. Substituting [tex]\(x = 1\)[/tex] into the original function, we encounter an undefined result since the denominator becomes zero. Therefore, the function is discontinuous at [tex]\(x = 1\)[/tex].
To confirm, we see that when substituting [tex]\( x = 1\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[
f(1) = \frac{3(1)^2 + 1 - 4}{1 - 1} = \frac{3 + 1 - 4}{0} = \frac{0}{0}
\][/tex]
This indicates an undefined value or a discontinuity at [tex]\(x = 1\)[/tex].
2. Find the Zeroes of the Function:
To find the zeroes of the function, we set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
3x^2 + x - 4 = 0
\][/tex]
Solving this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 3\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -4\)[/tex]:
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1 + 48}}{6}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{49}}{6}
\][/tex]
[tex]\[
x = \frac{-1 \pm 7}{6}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{-1 + 7}{6} = \frac{6}{6} = 1
\][/tex]
[tex]\[
x = \frac{-1 - 7}{6} = \frac{-8}{6} = -\frac{4}{3}
\][/tex]
Therefore, the zeroes of the function are [tex]\(x = 1\)[/tex] and [tex]\(x = -\frac{4}{3}\)[/tex].
3. Determine the Correct Pair:
Given the discontinuity at [tex]\(x = 1\)[/tex], we need to find the zero that corresponds to one of the given answer choices. We already see that [tex]\(x = -\frac{4}{3}\)[/tex] is one of the roots:
[tex]\[
\left(-\frac{4}{3}, 0\right)
\][/tex]
Given the detailed analysis, the correct answer pairs the discontinuity and zero correctly as:
Discontinuity at [tex]\((1, \text{undefined})\)[/tex] since the function is undefined at [tex]\(x = 1\)[/tex], zero at [tex]\(\left(-\frac{4}{3}, 0\right)\)[/tex].
Among the provided options, the correct one is:
Discontinuity at [tex]\((1,7)\)[/tex], zero at [tex]\(\left(-\frac{4}{3}, 0\right)\)[/tex].
[tex]\[ \boxed{\text{Discontinuity at } (1,7), \text{ zero at } \left(-\frac{4}{3}, 0\right)} \][/tex]
