termine the range values of $$ x $$ which the following functions are increasing $$ \begin{array}{llll} ext { (a) } y=x^{2} & ext { (b) } y=1+3 x-2 x^{2} & ext { (c) } y=x+\frac{1}{x} & ext { (d) } y=\frac{2 x-1}{x^{2}+2}\end{array} $$ $$ \begin{array}{ll} ext { n: }\end{array} $$

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termine the range values of $$ x $$ which the following functions are increasing $$ \begin{array}{llll}	ext { (a) } y=x^{2} & 	ext { (b) } y=1+3 x-2 x^{2} & 	ext { (c) } y=x+\frac{1}{x} & 	ext { (d) } y=\frac{2 x-1}{x^{2}+2}\end{array} $$ $$ \begin{array}{ll}	ext { n: }\end{array} $$

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Let's determine the intervals where each of the given functions is increasing by analyzing their derivatives. ### (a) $$ y = x^{2} $$ 1. **First Derivative:** $$ y' = \frac{dy}{dx} = 2x $$ 2. **Determine where $$ y' > 0 $$:** $$ 2x > 0 \implies x > 0 $$ **Conclusion:** The function $$ y = x^{2} $$ is **increasing** on the interval: $$ (0, \infty) $$ --- ### (b) $$ y = 1 + 3x - 2x^{2} $$ 1. **First Derivative:** $$ y' = \frac{dy}{dx} = 3 - 4x $$ 2. **Determine where $$ y' > 0 $$:** $$ 3 - 4x > 0 \implies 4x < 3 \implies x < \frac{3}{4} $$ **Conclusion:** The function $$ y = 1 + 3x - 2x^{2} $$ is **increasing** on the interval: $$ \left( -\infty, \frac{3}{4} \right) $$ --- ### (c) $$ y = x + \frac{1}{x} $$ 1. **Domain:** $$ x \neq 0 $$ 2. **First Derivative:** $$ y' = \frac{dy}{dx} = 1 - \frac{1}{x^{2}} $$ 3. **Determine where $$ y' > 0 $$:** $$ 1 - \frac{1}{x^{2}} > 0 \implies 1 > \frac{1}{x^{2}} \implies x^{2} > 1 \implies |x| > 1 $$ **Conclusion:** The function $$ y = x + \frac{1}{x} $$ is **increasing** on the intervals: $$ \left( -\infty, -1 \right) \cup \left( 1, \infty \right) $$ --- ### (d) $$ y = \frac{2x - 1}{x^{2} + 2} $$ 1. **First Derivative (Using Quotient Rule):** $$ y' = \frac{d}{dx} \left( \frac{2x - 1}{x^{2} + 2} \right) = \frac{2(x^{2} + 2) - (2x - 1)(2x)}{(x^{2} + 2)^{2}} $$ Simplifying the numerator: $$ 2x^{2} + 4 - (4x^{2} - 2x) = -2x^{2} + 2x + 4 $$ So, $$ y' = \frac{-2x^{2} + 2x + 4}{(x^{2} + 2)^{2}} $$ 2. **Determine where $$ y' > 0 $$:** $$ -2x^{2} + 2x + 4 > 0 \implies 2x^{2} - 2x - 4 < 0 $$ Solving the quadratic inequality: $$ 2x^{2} - 2x - 4 = 0 \implies x = \frac{2 \pm \sqrt{(−2)^{2} - 4 \cdot 2 \cdot (−4)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 32}}{4} = \frac{2 \pm 6}{4} $$ Thus, the roots are: $$ x = 2 \quad \text{and} \quad x = -1 $$ The quadratic expression $$ 2x^{2} - 2x - 4 $$ is less than zero between the roots: $$ -1 < x < 2 $$ **Conclusion:** The function $$ y = \frac{2x - 1}{x^{2} + 2} $$ is **increasing** on the interval: $$ (-1, 2) $$ --- ### Summary - **(a)** $$ y = x^{2} $$ is increasing for $$ x > 0 $$. - **(b)** $$ y = 1 + 3x - 2x^{2} $$ is increasing for $$ x < \frac{3}{4} $$. - **(c)** $$ y = x + \frac{1}{x} $$ is increasing for $$ x < -1 $$ and $$ x > 1 $$. - **(d)** $$ y = \frac{2x - 1}{x^{2} + 2} $$ is increasing for $$ -1 < x < 2 $$. The function $$ y = x^{2} $$ is increasing for $$ x > 0 $$. The function $$ y = 1 + 3x - 2x^{2} $$ is increasing for $$ x < \frac{3}{4} $$. The function $$ y = x + \frac{1}{x} $$ is increasing for $$ x < -1 $$ and $$ x > 1 $$. The function $$ y = \frac{2x - 1}{x^{2} + 2} $$ is increasing for $$ -1 < x < 2 $$.

termine the range values of which the following functions are increasing

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