91Ó°ÊÓ

Firstly, differentiate the given function \(h(x) = \frac{x^2}{x-1}\) with respect to x using the quotient rule. The quotient rule is given as: \(\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}\) Here, u = \(x^2\) and v = \(x - 1\). So, differentiate u with respect to x: \(u' = \frac{d}{dx}(x^2) = 2x\) And differentiate v with respect to x: \(v' = \frac{d}{dx}(x - 1) = 1\) Now, apply the quotient rule: \(h'(x) = \frac{(x - 1)(2x) - (x^2)(1)}{(x - 1)^2}\)

Now, simplify the expression of the first derivative: \(h'(x) = \frac{2x^2 - 2x - x^2}{(x - 1)^2}\) \(h'(x) = \frac{x^2 - 2x}{(x - 1)^2}\)

To find the critical points, we should equate the first derivative to zero and find when it is undefined. So, 1. Set the first derivative equal to zero: \(\frac{x^2 - 2x}{(x-1)^2} = 0\) Solve for x: \(x^2 - 2x = 0\) Factor out x: \(x(x - 2) = 0\) Thus, x = 0 and x = 2 are the critical points from the numerator. 2. Find where the first derivative is undefined: \((x - 1)^2 = 0\) Solve for x: x = 1 So, x = 1 is another critical point where the derivative is undefined.

Now, we have critical points x = 0, x = 1, and x = 2. We'll use the first derivative test to determine the intervals where the function is increasing and decreasing. For this, we'll test values in each of the intervals determined by the critical points. Intervals to test: - \(-∞ < x < 0\) - \(0 < x < 1\) - \(1 < x < 2\) - \(2 < x < +∞\) To perform the first derivative test, choose a representative value from each interval and plug it into the first derivative, \(h'(x)\): 1. Interval \(-∞ < x < 0\): Let x = -1 \(h'(-1) = \frac{(-1)^2 - 2(-1)}{(-1 - 1)^2} = \frac{3}{4}\) Since \(h'(-1) > 0\), the function is increasing in this interval. 2. Interval \(0 < x < 1\): Let x = 0.5 \(h'(0.5) = \frac{(0.5)^2 - 2(0.5)}{(0.5 - 1)^2} = -\frac{1}{4}\) Since \(h'(0.5) < 0\), the function is decreasing in this interval. 3. Interval \(1 < x < 2\): The first derivative is undefined at x = 1, so it's neither increasing nor decreasing. 4. Interval \(2 < x < +∞\): Let x = 3 \(h'(3) = \frac{(3)^2 - 2(3)}{(3 - 1)^2} = \frac{3}{4}\) Since \(h'(3) > 0\), the function is increasing in this interval.

Based on the first derivative test results, the function is: - Increasing on the intervals \((-∞, 0)\) and \((2, +∞)\) - Decreasing on the interval \((0, 1)\) - Neither increasing nor decreasing at x = 1.