Example Question - piecewise function
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Piecewise Function Analysis
<p>Para analizar la función dada, vamos a considerar los dos casos.</p> <p>Para \( x > 1 \):</p> <p>\( f(x) = \frac{x^3 - a^3}{x - a} \)</p> <p>Este se puede simplificar usando la factorización de la diferencia de cubos:</p> <p>\( f(x) = \frac{(x - a)(x^2 + ax + a^2)}{x - a} = x^2 + ax + a^2 \) (para \( x \neq a \)).</p> <p>Para \( x < 1 \):</p> <p>\( f(x) = \frac{3\sqrt{x} - 3\sqrt{a}}{x - a} \)</p> <p>Podemos factorizar el numerador de la misma manera si es necesario para un análisis adicional.</p> <p>La continuidad de la función puede revisarse evaluando el límite de ambas ramas en \( x = 1 \) y comparando los valores de \( f(1) \) en ambas partes si se define.</p>
Function Defined by Parts and Problem Solving
La imagen muestra una función definida por partes \( f(x) \) de la siguiente manera: \[ f(x) = \begin{cases} x^2 - 3x + 6 & \text{si } x \leq 2 \\ \frac{1}{x - 2} & \text{si } 2 < x < 5 \\ 3x - 4 & \text{si } x \geq 5 \end{cases} \] Para resolver un problema con esta función, debes primero determinar en qué intervalo o parte estás trabajando basado en el valor de \( x \) que estás considerando. Por ejemplo, si necesitas evaluar \( f(x) \) para un \( x \) específico, debes ver cuál de las tres condiciones se aplica para ese valor de \( x \) y usar la expresión correspondiente. Sin más información sobre lo que necesitas con la función, eso es todo lo que puedo explicarte por ahora. Si tienes un valor específico de \( x \) para evaluar o alguna otra pregunta relacionada con la función, por favor proporciona esa información adicional para poder asistirte más específicamente.
Finding Values for Continuity in a Piecewise Function
The image shows a mathematics question regarding a piecewise-defined function. The function \( f(x) \) is defined differently for different intervals of \( x \): \[ f(x) = \begin{cases} \log_3(3x + a) & , x < 1 \\ 3x & , x = 1 \\ \sqrt{x + b} & , x > 1 \end{cases} \] The question states that the function is continuous at \( x = 1 \) and asks for the values of \( a \) and \( b \). Continuity at \( x = 1 \) means that the left-hand limit as \( x \) approaches 1, the right-hand limit as \( x \) approaches 1, and the function's value at \( x = 1 \) must all be equal. Let's calculate these: 1. The left-hand limit as \( x \) approaches 1 is the limit of \( \log_3(3x + a) \) as \( x \) approaches 1 from the left: \[ \lim_{{x \to 1^-}} \log_3(3x + a) = \log_3(3 \cdot 1 + a) = \log_3(3 + a) \] 2. The function's value at \( x = 1 \) is given directly as \( f(1) = 3 \cdot 1 = 3 \). 3. The right-hand limit as \( x \) approaches 1 is the limit of \( \sqrt{x + b} \) as \( x \) approaches 1 from the right: \[ \lim_{{x \to 1^+}} \sqrt{x + b} = \sqrt{1 + b} \] For continuity at \( x = 1 \), we must have \( \log_3(3 + a) = 3 = \sqrt{1 + b} \). From here we can find the values of \( a \) and \( b \): \(\log_3(3 + a) = 3\) implies \(3^3 = 3 + a\), hence \( a = 27 - 3 = 24 \). \(\sqrt{1 + b} = 3\) implies \(1 + b = 3^2\), hence \( b = 9 - 1 = 8 \). Thus, \( a = 24 \) and \( b = 8 \).
Determining Values for Function Continuity
The image shows a piecewise function with three different expressions for different intervals of x: 1. \(f(x) = \log_2(3x + a)\) for \(x < 1\) 2. \(f(x) = 3^x\) for \(x = 1\) 3. \(f(x) = \sqrt{x + b}\) for \(x > 1\) The question asks for the values of \(a\) and \(b\) such that the function \(f(x)\) is continuous at \(x = 1\). To ensure continuity at \(x = 1\), the following conditions must be met: 1. The limit of \(f(x)\) as \(x\) approaches 1 from the left must equal the value of \(f(1)\). 2. The limit of \(f(x)\) as \(x\) approaches 1 from the right must also equal the value of \(f(1)\). Let’s find these limits: 1. The limit as \(x\) approaches 1 from the left: \[ \lim_{{x \to 1^-}} \log_2(3x + a) = \log_2(3 \cdot 1 + a) = \log_2(3 + a) \] 2. The limit as \(x\) approaches 1 from the right: \[ \lim_{{x \to 1^+}} \sqrt{x + b} = \sqrt{1 + b} \] We also know that \(f(1) = 3^1 = 3\). Setting these limits equal to \(f(1)\), we get: \[ \log_2(3 + a) = 3 \] \[ \sqrt{1 + b} = 3 \] From the first equation, we can solve for \(a\): \[ 2^3 = 3 + a \] \[ 8 = 3 + a \] \[ a = 5 \] From the second equation, we can solve for \(b\): \[ 3^2 = 1 + b \] \[ 9 = 1 + b \] \[ b = 8 \] Therefore, the values of \(a\) and \(b\) are 5 and 8, respectively.
Piecewise Graph Limit Calculation
The image presents a piecewise graph of the function y = f(x) and asks for the value of the expression: \[ \lim_{x \to -4^+} f(x) - \lim_{x \to 0^-} f(x) + \lim_{x \to 2^+} f(x) \] From the graph, we can derive the following limits: 1. \( \lim_{x \to -4^+} f(x) \) is the value that f(x) approaches as x approaches -4 from the right. By looking at the graph, we can see this value is n. 2. \( \lim_{x \to 0^-} f(x) \) is the value that f(x) approaches as x approaches 0 from the left. On the graph, this value is m. 3. \( \lim_{x \to 2^+} f(x) \) is the value that f(x) approaches as x approaches 2 from the right. The graph shows this value as k. Now, we substitute these values into the given expression: n (the limit as x approaches -4 from the right) - m (the limit as x approaches 0 from the left) + k (the limit as x approaches 2 from the right). From the graph, we can infer the following based on the labelled points: - n is at y = -1 - m is at y = 1 - k is at y = 0 Using these values: \( \lim_{x \to -4^+} f(x) = -1 \) \( \lim_{x \to 0^-} f(x) = 1 \) \( \lim_{x \to 2^+} f(x) = 0 \) Substitute these values into the expression: \[ (-1) - (1) + (0) = -2 \] Therefore, the value of the given expression is -2.
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