Example Question - reciprocal
Here are examples of questions we've helped users solve.
Exponentiation of Fractions with Negative Exponent
Para resolver la expresión \(\left(\frac{16}{81}\right)^{-\frac{5}{4}}\), sigue estos pasos: 1. Primero, recuerda que elevar un número a un exponente negativo es equivalente a tomar el recíproco del número y cambiar el signo del exponente a positivo. Es decir, \(a^{-b} = \frac{1}{a^b}\). Entonces, la expresión se convierte en: \(\left(\frac{16}{81}\right)^{-\frac{5}{4}} = \frac{1}{\left(\frac{16}{81}\right)^{\frac{5}{4}}}\) 2. Ahora vamos a trabajar con la parte del exponente \(\left(\frac{16}{81}\right)^{\frac{5}{4}}\). Para elevar una fracción a una potencia fraccionaria, elevamos el numerador y el denominador por separado: \(\left(\frac{16}{81}\right)^{\frac{5}{4}} = \left(16^{\frac{5}{4}}\right) / \left(81^{\frac{5}{4}}\right)\) 3. Las potencias fraccionarias pueden descomponerse en una raíz y una potencia. Por ejemplo, \(a^{\frac{m}{n}}\) es lo mismo que \(\sqrt[n]{a^m}\). En este caso, \(a = 16\) y \(a = 81\), \(m = 5\) y \(n = 4\). \(\left(16^{\frac{5}{4}}\right) / \left(81^{\frac{5}{4}}\right) = \left(\sqrt[4]{16^5}\right) / \left(\sqrt[4]{81^5}\right)\) 4. Sabemos que \(\sqrt[4]{16} = 2\) y \(\sqrt[4]{81} = 3\), porque \(2^4 = 16\) y \(3^4 = 81\). Entonces, elevamos 2 y 3 a la quinta potencia: \(\left(\sqrt[4]{16^5}\right) / \left(\sqrt[4]{81^5}\right) = \left(2^5\right) / \left(3^5\right)\) 5. Calculamos \(2^5 = 32\) y \(3^5 = 243\). \(\left(2^5\right) / \left(3^5\right) = \frac{32}{243}\) 6. Finalmente, volvemos a la expresión original donde teníamos el recíproco: \(\frac{1}{\left(\frac{16}{81}\right)^{\frac{5}{4}}} = \frac{1}{\frac{32}{243}}\) 7. Tomamos el recíproco de \(\frac{32}{243}\), que es \(\frac{243}{32}\). Por lo tanto, el resultado de la expresión original \(\left(\frac{16}{81}\right)^{-\frac{5}{4}}\) es \(\frac{243}{32}\).
Calculating the Inverse of a Derivative at a Specific Point
To solve for the value of \((df/dx)^{-1}\) at the point \(x = 398 = f(5)\), we are essentially looking to compute the reciprocal of the derivative of \(f(x)\) at \(x = 5\). The reciprocal of the derivative is also known as the derivative of the inverse function evaluated at the corresponding y-value, which in this case is 398. First, we'll find \(df/dx\) by differentiating \(f(x)\): \(f(x) = 5x^3 - 9x^2 - 2\) \(df/dx = 15x^2 - 18x\) Now we need to evaluate \(df/dx\) at \(x = 5\): \(df/dx|_{x=5} = 15(5)^2 - 18(5) = 15(25) - 90 = 375 - 90 = 285\) Now, \((df/dx)^{-1}\) at \(x = 398\) is the reciprocal of this value: \((df/dx)^{-1}|_{f(5)} = 1 / df/dx|_{x=5} = 1/285\) So, the value is \(\frac{1}{285}\), which is the simplified fraction we were looking to find.
Solving Expression with Reciprocal
To solve the expression x + \(\frac{1}{x}\) given that \(x = 2 + \sqrt{3}\), first find the reciprocal of x and then add it to x. Given \(x = 2 + \sqrt{3}\), the reciprocal, \(\frac{1}{x}\), can be calculated as follows: \[ \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \] To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator: \[ \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} \] Now, apply the difference of squares to the denominator: \[ \frac{1}{x} = \frac{2 - \sqrt{3}}{4 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} \] Now, add x to \(\frac{1}{x}\): \[ x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) \] When you combine the terms, the \(\sqrt{3}\) terms will cancel out: \[ x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} = 2 + 2 = 4 \] So, \(x + \frac{1}{x}\) is equal to 4.
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