Example Question - rational expression
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Solving an Inequality Involving a Rational Expression
<p>\frac{3}{1 - x} \leq 2</p> <p>3 \leq 2(1 - x)</p> <p>3 \leq 2 - 2x</p> <p>2x \leq 2 - 3</p> <p>2x \leq -1</p> <p>x \geq \frac{-1}{2}</p> <p>Karena untuk \( x<1 \), penyebut \( 1-x \) adalah positif dan karena itu tidak mengubah ketidaksamaan saat kita mengalikannya. Namun, kita harus mempertimbangkan apa yang terjadi saat \( x>1 \).</p> <p>Jika \( x>1 \), maka penyebut \( 1-x \) negatif, dan ketidaksamaan harus dibalik ketika kita mengalikannya:</p> <p>\frac{3}{1 - x} \leq 2, harus diubah menjadi \frac{3}{1 - x} \geq 2</p> <p>Ini tidak memenuhi kondisi awal \( \frac{3}{1 - x} \leq 2 \), jadi tidak ada solusi di \( x>1 \).</p> <p>Oleh karena itu, solusi untuk ketidaksamaan ini adalah \( x \geq \frac{-1}{2} \) dan \( x<1 \).</p> <p>Interval solusi dalam notasi interval adalah \( \left[\frac{-1}{2}, 1\right) \).</p>
Simplification of a Rational Expression
<p>To simplify the given expression, he can factor the denominator and check if there are common factors between numerator and denominator that he can cancel out:</p> <p>The denominator is an expression that looks like the difference of squares, which can be factored as:</p> <p>\((2x-1)^2 - 3(x^2-1) = (2x-1)^2 - 3(x+1)(x-1)\)</p> <p>He should expand the squares and multiply out the terms to simplify:</p> <p>\((2x-1)^2 = (2x-1)(2x-1) = 4x^2 - 4x + 1\)</p> <p>\(3(x+1)(x-1) = 3(x^2 - 1) = 3x^2 - 3\)</p> <p>Subtracting the second expression from the first:</p> <p>\(4x^2 - 4x + 1 - (3x^2 - 3) = 4x^2 - 4x + 1 - 3x^2 + 3\)</p> <p>Simplifying:</p> <p>\(x^2 - 4x + 4\)</p> <p>He then recognizes this as a perfect square trinomial which can be factored as:</p> <p>\(x^2 - 4x + 4 = (x-2)^2\)</p> <p>So the expression simplifies to:</p> <p>\(\frac{x}{(x-2)^2}\)</p> <p>He notes that there are no common factors to cancel out with the numerator, hence this is the simplified form of the given rational expression.</p>
Simplifying a Rational Expression Involving Polynomials
<p> \( y = \frac{x}{(2x-1) \cdot (3) \cdot (x-1)} \) </p> <p> To simplify the expression, factor out and reduce common terms if any. </p> <p> No common terms to factor out or reduce. </p> <p> The expression is already in its simplest form. So, the simplified expression is: </p> <p> \( y = \frac{x}{6x^2 - 3x - 2x + 1} \) </p> <p> \( y = \frac{x}{6x^2 - 5x + 1} \) </p>
Graphical Representation of a Rational Inequality Solution
<p>To find the solution to the inequality \((3x + 1)^2 > 3(3x + 1)\), we start by expanding and simplifying:</p> <p>\((3x + 1)(3x + 1) > 3(3x + 1)\)</p> <p>\(9x^2 + 6x + 1 > 9x + 3\)</p> <p>\(9x^2 + 6x - 9x + 1 - 3 > 0\)</p> <p>\(9x^2 - 3x - 2 > 0\)</p> <p>We then factor the quadratic expression, if possible, or use the quadratic formula to find the roots (the values of \(x\) where \(9x^2 - 3x - 2 = 0\)).</p> <p>Factoring gives us:</p> <p>\((3x + 1)(3x - 2) > 0\)</p> <p>The roots are \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\). The inequality will change signs at each of these points. To determine where the expression is positive, we can test intervals that are defined by these roots. We choose test points from intervals \((-∞, -\frac{1}{3})\), \((- \frac{1}{3}, \frac{2}{3})\), and \((\frac{2}{3}, ∞)\).</p> <p>Upon testing, we find the inequality is satisfied for \(x < -\frac{1}{3}\) and \(x > \frac{2}{3}\).</p> <p>On the number line, the solution will be shown with an open circle at \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\) and shading to the left of \(x = -\frac{1}{3}\) and to the right of \(x = \frac{2}{3}\), because these are not included in the solution set.</p> <p>Therefore, the correct answer is option D.</p>
Finding the Limit of a Rational Expression
<p>Para resolver el límite dado, sustituimos directamente el valor al cual tiende la variable:</p> <p>\[\lim_{{t \to 3}} \frac{{t - 2}}{{t + 5}} = \frac{{3 - 2}}{{3 + 5}}\]</p> <p>Realizamos las operaciones dentro del límite:</p> <p>\[\frac{{1}}{{8}}\]</p> <p>Por lo tanto, el resultado del límite es \(\frac{{1}}{{8}}\).</p>
Derivative of a Function with Rational Expression
Given the function: \( h(x) = \frac{4x^3 - 7x + 8}{x} \) First, simplify \( h(x) \) by dividing each term in the numerator by \( x \): \( h(x) = 4x^2 - 7 + \frac{8}{x} \) Now, find the derivative \( h'(x) \): \( h'(x) = \frac{d}{dx}(4x^2) - \frac{d}{dx}(7) + \frac{d}{dx}\left(\frac{8}{x}\right) \) \( h'(x) = 8x - 0 - 8x^{-2} \) \( h'(x) = 8x - 8x^{-2} \) Expressing \( -8x^{-2} \) as \( -\frac{8}{x^2} \), we have: \( h'(x) = 8x - \frac{8}{x^2} \)
Derivative of a Function with a Rational Expression
To find the derivative of the function \( h(x) \) given by: \[ h(x) = \frac{4x^3 - 7x + 8}{x} \] you can simplify the function before differentiating it by dividing each term in the numerator by \( x \): \[ h(x) = \frac{4x^3}{x} - \frac{7x}{x} + \frac{8}{x} \] \[ h(x) = 4x^2 - 7 + \frac{8}{x} \] Now, we'll differentiate each term separately: - The derivative of \( 4x^2 \) with respect to \( x \) is \( 8x \). - The derivative of a constant, \( -7 \), is 0. - The derivative of \( \frac{8}{x} \), which is \( 8x^{-1} \), is \( -8x^{-2} \) or \( -\frac{8}{x^2} \). So, the derivative of \( h(x) \), which we denote as \( h'(x) \), is: \[ h'(x) = 8x - 0 - \frac{8}{x^2} \] \[ h'(x) = 8x - \frac{8}{x^2} \] This is the final form of the derivative of \( h(x) \).
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