Example Question - logarithmic function
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Finding Domain of Logarithmic Function
Claro, abordemos la pregunta. Se nos pide encontrar el dominio de la función \(f(x) = \log\left(\frac{1+x}{1-x}\right)\). El dominio de una función es el conjunto de todos los valores de entrada (x) para los cuales la función está definida. En este caso, estamos tratando con una función logarítmica. Los logaritmos están definidos sólo para argumentos positivos, lo que significa que el argumento del logaritmo \(\left(\frac{1+x}{1-x}\right)\) debe ser mayor que cero. Establezcamos esta condición: \[\frac{1+x}{1-x} > 0\] Para resolver esta desigualdad, primero hay que encontrar los valores que hacen que el numerador y el denominador sean cero, ya que estos son los valores que no están en el dominio (puntos de discontinuidad). Para el numerador tenemos \(1+x = 0\), lo que nos da \(x = -1\). Para el denominador tenemos \(1-x = 0\), lo que nos da \(x = 1\). Estos valores dividen la línea de números reales en intervalos. Debemos probar cada intervalo para ver si satisface la desigualdad: Los intervalos son: \((-∞, -1)\), \((-1, 1)\) y \((1, ∞)\). 1. Para un número en el intervalo \((-∞, -1)\), como \(x = -2\), la expresión se convierte en un número positivo dividido entre otro positivo, lo que resulta en un valor positivo que cumple la desigualdad. Por lo tanto, todo el intervalo \((-∞, -1)\) está en el dominio. 2. Para un número en el intervalo \((-1, 1)\), como \(x = 0\), obtenemos \(1 > 0\), lo que nuevamente cumple la desigualdad. Así, todo el intervalo \((-1, 1)\) está en el dominio. 3. Para un número en el intervalo \((1, ∞)\), como \(x = 2\), la expresión se convierte en negativo dividido entre negativo, lo que da un positivo. Por lo que este intervalo también cumple la desigualdad y está en el dominio. Sin embargo, debemos ser cuidadosos de excluir los valores \(x = -1\) y \(x = 1\) del dominio ya que estos hacen que el denominador sea cero, lo que haría que la expresión no esté definida (no podemos tener una división por cero). Por lo tanto, el dominio de \(f(x)\) es todos los números reales excepto \(x = -1\) y \(x = 1\), lo que matemáticamente podríamos escribir como: \[Dom(f) = (-∞, -1) ∪ (-1, 1) ∪ (1, ∞)\]
Partial Derivative Operations of Logarithmic Function
题目要求对函数 \( z = \ln(xy) \) 进行偏导运算。 首先，找到 \( \frac{\partial^3 z}{\partial x^2 \partial y} \) 和 \( \frac{\partial^3 z}{\partial x \partial y^2} \)，然后计算它们的差值。 函数 \( z \) 关于 \( x \) 的一阶偏导数是： \[ \frac{\partial z}{\partial x} = \frac{1}{xy} \cdot y = \frac{1}{x} \] 函数 \( z \) 关于 \( y \) 的一阶偏导数是： \[ \frac{\partial z}{\partial y} = \frac{1}{xy} \cdot x = \frac{1}{y} \] 接下来，计算它们的二阶偏导数和三阶偏导数。 对 \( \frac{\partial z}{\partial x} \) 关于 \( x \) 求二阶偏导数： \[ \frac{\partial^2 z}{\partial x^2} = -\frac{1}{x^2} \] 对 \( \frac{\partial^2 z}{\partial x^2} \) 关于 \( y \) 求三阶偏导数： \[ \frac{\partial^3 z}{\partial x^2 \partial y} = 0 \] 对 \( \frac{\partial z}{\partial y} \) 关于 \( y \) 求二阶偏导数： \[ \frac{\partial^2 z}{\partial y^2} = -\frac{1}{y^2} \] 对 \( \frac{\partial^2 z}{\partial y^2} \) 关于 \( x \) 求三阶偏导数： \[ \frac{\partial^3 z}{\partial x \partial y^2} = 0 \] 由于 \( \frac{\partial^3 z}{\partial x^2 \partial y} \) 和 \( \frac{\partial^3 z}{\partial x \partial y^2} \) 都等于 0，所以： \[ \frac{\partial^3 z}{\partial x^2 \partial y} - \frac{\partial^3 z}{\partial x \partial y^2} = 0 - 0 = 0 \] 因此，所求的差值是 0。
Partial Derivatives of a Logarithmic Function
为了求解这个偏微分方程，我们首先需要计算出 \(\frac{\partial^3 z}{\partial x^2 \partial y}\) 和 \(\frac{\partial^3 z}{\partial x \partial y^2}\)。 给定函数 \(z = x \ln(y)\)，我们首先对 \(x\) 和 \(y\) 进行偏微分。 第一步，计算 \(z\) 对 \(x\) 的一阶偏导数： \[\frac{\partial z}{\partial x} = \ln(y) + x \cdot \frac{1}{y} \cdot 0 = \ln(y)\] 第二步，计算 \(z\) 对 \(y\) 的一阶偏导数： \[\frac{\partial z}{\partial y} = x \cdot \frac{1}{y}\] 第三步，求第一步结果对 \(x\) 的二阶偏导数： \[\frac{\partial^2 z}{\partial x^2} = 0\] 第四步，求第二步结果对 \(y\) 的二阶偏导数： \[\frac{\partial^2 z}{\partial y^2} = -x \cdot \frac{1}{y^2}\] 第五步，求第二步结果对 \(x\) 的混合二阶偏导数： \[\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left( x \cdot \frac{1}{y} \right) = \frac{1}{y}\] 第六步，计算对 \(x\) 的二阶偏导数和 \(y\) 的一阶偏导数的混合三阶偏导数： \[\frac{\partial^3 z}{\partial x^2 \partial y} = \frac{\partial}{\partial y} \left( \frac{\partial^2 z}{\partial x^2} \right) = \frac{\partial}{\partial y}(0) = 0\] 第七步，计算对 \(x\) 的一阶偏导数和 \(y\) 的二阶偏导数的混合三阶偏导数： \[\frac{\partial^3 z}{\partial x \partial y^2} = \frac{\partial}{\partial x} \left( \frac{\partial^2 z}{\partial y^2} \right) = \frac{\partial}{\partial x} \left( -x \cdot \frac{1}{y^2} \right) = -\frac{1}{y^2}\] 所以，我们得到： \[\frac{\partial^3 z}{\partial x^2 \partial y} = 0\] \[\frac{\partial^3 z}{\partial x \partial y^2} = -\frac{1}{y^2}\] 我们现在可以将两个结果代入方程计算最终的答案： \[\frac{2024}{0} - \frac{0}{-\frac{1}{y^2}} = \infty\] 但这个除法没有定义，因为除数是零。因此，给出的方程没有数学意义。
Logarithmic Function Transformation Analysis
The logarithmic function \( f(x) = \ln(x) \) is transformed into \( g(x) = \ln(x - 2) + 5 \). To determine which of the provided statements is true, we need to analyze the transformation in two parts: the horizontal shift caused by the change inside the logarithm and the vertical shift caused by the addition outside the logarithm. The \( \ln(x) \) function is shifted horizontally by the presence of \(-2\) inside the function, which means the function is shifted to the right by 2 units. That's because the logarithm’s argument \( x \) is replaced with \( x - 2 \), indicating a rightward shift of 2 units. On the other hand, adding \( +5 \) to the function after the logarithm \( \ln(x - 2) \) indicates a vertical shift. Specifically, it means the graph of the function is shifted upwards by 5 units. Given these transformations: A. Incorrect: \( f(x) \) is not translated up 5 units; it's the function \( g(x) \) that has been translated up by 5 units compared to \( f(x) \). B. Incorrect: \( f(x) \) is not translated down 2 units. C. Incorrect: \( f(x) \) is not translated left 2 units; it's moving to the right. D. Correct: \( f(x) \) is translated right 5 units, which exactly describes the horizontal shift. Therefore, the correct answer is D: \( f(x) \) is translated right 5 units.
Logarithmic Function Transformations
The given logarithmic function f(x) = ln x is transformed to g(x) = ln(x - 2) + 5. Comparing the function g(x) with f(x): 1. The "+ 5" at the end of g(x) indicates a vertical translation upward by 5 units. 2. The "(x - 2)" inside the logarithm in g(x) suggests a horizontal translation to the right by 2 units since the effect of subtracting inside the function argument is to shift the graph to the right. Given the options provided: A. f(x) is translated up 5 units. (This is correct for the vertical translation.) B. f(x) is translated down 2 units. (This is incorrect; the function is not translated downward.) C. f(x) is translated left 2 units. (This is incorrect; the function is actually translated right.) D. f(x) is translated right 5 units. (This is incorrect; the translation is 2 units to the right, not 5.) Therefore, the correct answer is A. f(x) is translated up 5 units.
Logarithmic Function Transformation Analysis
The given logarithmic function f(x) = log(x) is transformed to g(x) = log(x + 1) + 3. To determine which of the statements is true, we need to analyze the transformation that g(x) represents compared to f(x). - For statement A: f(x) is translated 1 unit upward. This is not true because the function g(x) has a "+3" outside of the logarithm, which means the entire function is translated 3 units upward, not 1 unit. - For statement B: f(x) is translated 3 units downward. This is incorrect as the transformation involves a "+3", indicating an upward translation, not a downward one. - For statement C: The vertical asymptote shifts 1 unit to the left. This is true. The vertical asymptote for the basic logarithmic function f(x) = log(x) is at x=0. Since g(x) is log(x + 1), this transformation shifts the graph horizontally 1 unit to the left, meaning the new vertical asymptote is now at x=-1. - For statement D: The vertical asymptote shifts 3 units to the right. This is not true. The transformation inside the logarithmic function (x + 1) does not shift the graph to the right; it shifts it to the left. Hence, the correct statement is C: The vertical asymptote shifts 1 unit to the left.
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