Example Question - domain
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Graphing Intervals and Functions
<p>Para representar gráficamente los intervalos, comenzamos con:</p> <p>a. [1, 2): Una línea sólida desde 1 hasta 2, sin incluir 2.</p> <p>b. (-∞, 2]: Una línea sólida que se extiende desde -∞ hasta 2, incluyendo 2.</p> <p>c. (-5, 0): Una línea sólida desde -5 hasta 0, sin incluir 0.</p> <p>d. [2, ∞): Una línea sólida desde 2 hacia el infinito, incluyendo 2.</p> <p>e. (-∞, -1]: Una línea sólida que se extiende desde -∞ hasta -1, incluyendo -1.</p> <p>Para la función:</p> <p>a. \( y = \sqrt{x - 4} \): El dominio es \( x \geq 4 \) y el rango es \( y \geq 0 \).</p> <p>b. \( y = \sqrt{x + 1} \): El dominio es \( x \geq -1 \) y el rango es \( y \geq 0 \).</p>
Finding the Domain and Range of a Given Function
<p>The given function is \( f(x) = \frac{1}{\sqrt{9-x^2}} \).</p> <p>To find the domain, we need to identify all values of \( x \) for which the function is defined. The denominator cannot be zero, and the expression under the square root must be non-negative because the root must be a real number.</p> <p>We have the inequality \( 9 - x^2 \geq 0 \) which leads to \( -3 \leq x \leq 3 \).</p> <p>Hence, the domain of \( f(x) \) is \( [-3, 3] \).</p> <p>To find the range, we know that the function is always positive because it is the reciprocal of the square root of a positive number, and the square root function only returns non-negative values.</p> <p>Since the square root function \( \sqrt{9-x^2} \) will have values ranging from \( 0 \) (not included because it would make the denominator zero) to \( 3 \), the reciprocal of this function will range from \( \frac{1}{3} \) to infinity.</p> <p>Hence, the range of \( f(x) \) is \( (0, \infty) \).</p>
Domain and Range Determination for Given Functions
<p>Untuk soal (a):</p> <p>D_{f(x,y)} = \{(x, y) \in \mathbb{R}^2 | x^2 - y^2 - 9 \geq 0\}</p> <p>R_{f(x,y)} = \{f(x,y) \in \mathbb{R} | f(x,y) = 4-x^2-y^2 + \frac{1}{2}\log(x^2 - y^2 - 9), (x, y) \in D_{f(x,y)} \}</p> <p>Untuk soal (b):</p> <p>D_{f(x,y)} = \{(x, y) \in \mathbb{R}^2 | x^2 + y^2 - 1 \neq 0\}</p> <p>R_{f(x,y)} = \{f(x,y) \in \mathbb{R} | f(x,y) = \frac{\sqrt{x^2 - y^2}}{x^2 + y^2 - 1}, (x, y) \in D_{f(x,y)} \}</p>
Analysis of Parabola Characteristics
La imagen proporciona dos funciones cuadráticas para analizar sus características. Vamos a determinar estas características para la primera función \( f(x) = 2x^2 - 6x - 4 \). <p>Para encontrar la orientación de la parábola, observamos el coeficiente líder \(a\).</p> <p>Si \( a > 0 \), la parábola se abre hacia arriba. Si \( a < 0 \), se abre hacia abajo.</p> <p>En nuestro caso, \( a = 2 \), por lo que la parábola se abre hacia arriba.</p> <p>El eje de simetría de una parábola se encuentra en \( x = -\frac{b}{2a} \).</p> <p>Sustituimos \( a = 2 \), \( b = -6 \) para obtener \( x = -\frac{-6}{2 \cdot 2} = \frac{3}{2} \).</p> <p>El vértice de la parábola se encuentra en el punto \( (\frac{-b}{2a}, f(\frac{-b}{2a})) \).</p> <p>Calculamos \( f(\frac{3}{2}) = 2(\frac{3}{2})^2 - 6(\frac{3}{2}) - 4 \).</p> <p>El vértice es \( (\frac{3}{2}, -\frac{25}{4}) \).</p> <p>El intercepto en x son los ceros de la función, donde \( f(x) = 0 \).</p> <p>Resolvemos \( 2x^2 - 6x - 4 = 0 \) usando la fórmula cuadrática o factorización para encontrar los ceros.</p> <p>El intercepto en y es \( f(0) \), es decir, \( -4 \).</p> <p>El dominio de cualquier función cuadrática es \( (-\infty, \infty) \).</p> <p>El recorrido (rango) depende de la orientación de la parábola. Como se abre hacia arriba, el rango es \( [f(\frac{-b}{2a}), \infty) \), es decir, \( [-\frac{25}{4}, \infty) \).</p> Este es un análisis completo para la primera función. Para la segunda función, se seguiría un proceso similar.
Determining the Domain and Range from a Graph
<p>The graph shows a function with two distinct parts. The first part is decreasing and the second part is increasing. There is a break in the graph where the function is not defined.</p> <p>To find the domain, we look for the x-values that the function covers. By observing the graph, we see that the function is defined for all x except for a portion where x is between -4 and 3. Thus the domain is \( x < -4 \) or \( x > 3 \).</p> <p>The range of a function is the set of all possible output values (y-values), which result from using the function's formula. By examining the graph, we see that as \( x \) approaches -4 from the left, the y-values decrease without bound, and as \( x \) approaches 3 from the right, the y-values increase without bound. Therefore, the range of the function is all real numbers, which can be denoted as \( -\infty < y < \infty \) or simply \( y \in \mathbb{R} \).</p>
Determining the Domain and Range from a Graph
\[ \text{Domain: } \{ x \mid x < -4 \text{ or } x \geq 4\} \] \[ \text{Range: } \{ y \mid y \geq 4 \} \]
Finding the Inverse of a Quadratic Function
<p>Sea \( g(x) = x^2 - 1 \) la función dada, y queremos encontrar \( g^{-1}(x) \).</p> <p>Para encontrar la función inversa, primero reemplazamos \( g(x) \) por \( y \):</p> <p>\( y = x^2 - 1 \)</p> <p>Luego, resolvemos para \( x \) en términos de \( y \):</p> <p>\( y + 1 = x^2 \)</p> <p>\( x = \sqrt{y + 1} \), pero dado que el dominio está restringido a los números reales positivos, asumimos que \( x \) también es positivo.</p> <p>Ahora intercambiamos \( x \) e \( y \) para obtener la función inversa:</p> <p>\( y = \sqrt{x + 1} \)</p> <p>Por lo tanto, la expresión algebraica de \( g^{-1}(x) \) es \( \sqrt{x + 1} \).</p>
Understanding Functions and Their Inverses
The image displays two functions, h and g. The function h is defined as h: x → 2x+1, for real numbers x such that 0 ≤ x ≤ k, and the function g is defined as g: x → 2x for 0 ≤ x ≤ 1/2 and 2 - 2x for 1/2 ≤ x ≤ 1. Part i of the question asks: Given that the function h has an inverse, state the greatest value of k. Find h⁻¹(½) and g(h⁻¹(½)). To solve part i, let's break it down: 1. **Function h having an inverse:** A function has an inverse if and only if it is one-to-one (injective), which means it never assigns the same value to two different domain elements. The function h given by h(x) = 2x + 1 is a linear function with a positive slope, which is injective over the entire domain of real numbers. Thus, there is no restriction on the value of k for h to have an inverse, and the greatest value of k is the largest value in the domain of x given for h, which is ∞. However, given the typical constraints of such problems, the greatest value of k could be considered the maximum value for which h is defined, which would be k if we associate it with the boundary of the domain. Since the problem does not specify an upper limit to the domain, we assume k can grow indefinitely. 2. **Finding h⁻¹(½):** The inverse function, h⁻¹(x), will undo the operation performed by h(x). That is, if h(x) = 2x + 1, then for h⁻¹(y), where y = h(x), we must have x = (y - 1)/2. So, h⁻¹(½) means we should solve (½ - 1)/2 for x: h⁻¹(½) = (½ - 1)/2 = (-½)/2 = -¼ 3. **Finding g(h⁻¹(½)):** Now that we have x = -¼, we need to determine which piece of the piecewise function g to use. Since -¼ is not within the domain of either piece of g as defined, g(h⁻¹(½)) is not directly determined from the given definition of g and this seems to be an issue with either the domain of g as presented or the parameters of the problem. Typically, with the domain given for g, it's likely intended for positive values of x, and h⁻¹(½) gives us a negative value. Thus, there must be an assumption or additional information about function g beyond the given domain, or an error in the problem as written. In conclusion, for part i, the greatest value of k isn't constrained by the information given, and to find g(h⁻¹(½)), we would need further clarification on the behavior of g beyond the domain provided in the image.
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