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$g(x)= \sqrt[3]{x-1}$

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Solve quadratic inequalities

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Q.

g(x)= \sqrt[3]{x-1}

Find Zeros: Find the zeros of the quadratic expression by setting $(x - 1)(x + 1) = 0$ . $x - 1 = 0$ gives $x = 1$ . $x + 1 = 0$ gives $x = -1$ .Critical points are $x = 1$ and $x = -1$ .
Determine Sign: Determine the sign of $(x - 1)(x + 1)$ in each interval created by the critical points: $(-\infty, -1)$ , $(-1, 1)$ , and $(1, \infty)$ .
Test Intervals: Test a point in the interval $(-\infty, -1)$ , say $x = -2$ . $(x - 1)(x + 1)$ at $x = -2$ is $(-2 - 1)(-2 + 1) = (-3)(-1) = 3$ , which is positive.
Combine Intervals: Test a point in the interval $(-1, 1)$ , say $x = 0$ . $\newline$ $(x - 1)(x + 1) \ at \ x = 0 \ is \ (0 - 1)(0 + 1) = (-1)(1) = -1$ , which is negative.
Combine Intervals: Test a point in the interval $(-1, 1)$ , say $x = 0$ . $(x - 1)(x + 1)$ at $x = 0$ is $(0 - 1)(0 + 1) = (-1)(1) = -1$ , which is negative.Test a point in the interval $(1, \infty)$ , say $x = 2$ . $(x - 1)(x + 1)$ at $x = 2$ is $(2 - 1)(2 + 1) = (1)(3) = 3$ , which is positive.
Combine Intervals: Test a point in the interval $(-1, 1)$ , say $x = 0$ .
$(x - 1)(x + 1)$ at $x = 0$ is $(0 - 1)(0 + 1) = (-1)(1) = -1$ , which is negative.Test a point in the interval $(1, \infty)$ , say $x = 2$ .
$(x - 1)(x + 1)$ at $x = 2$ is $(2 - 1)(2 + 1) = (1)(3) = 3$ , which is positive.Combine the intervals where $(x - 1)(x + 1)$ is positive.
The solution is $x = 0$ $1$ or $x = 0$ $2$ .

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