Example Question - domain and range
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Multiple Mathematics Problems Involving Algebra, Geometry, and Complex Numbers
<p>First question: Find the 4th term from the end in the expansion of \( \left( \frac{3}{x^2} - x^3 \right)^7 \).</p> <p>The 4th term from the end is the same as the 4th term from the beginning, which corresponds to \( T_4 \) in the expansion:</p> <p>\( T_k = \binom{n}{k-1} \cdot (a)^{n-(k-1)} \cdot (b)^{k-1} \) where \( n = 7 \), \( a = \frac{3}{x^2} \), and \( b = -x^3 \).</p> <p>\( T_4 = \binom{7}{3} \cdot \left(\frac{3}{x^2}\right)^{7-3} \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \left(\frac{3}{x^2}\right)^4 \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \frac{81}{x^8} \cdot (-x^9) \)</p> <p>\( T_4 = 35 \cdot 81 \cdot \frac{-x^9}{x^8} \)</p> <p>\( T_4 = -2835 \cdot \frac{1}{x} \)</p> <p>Second question: Find the equation of lines passing through (1,2) and making angle 30° with y-axis.</p> <p>The slope of the line making a 30° angle with the y-axis is the tangent of (90°-30°), which is \( \tan(60°) \).</p> <p>Slope (m) of the desired line: \( m = \tan(60°) = \sqrt{3} \)</p> <p>Use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equation of the line passing through (1,2).</p> <p>\( y - 2 = \sqrt{3}(x - 1) \)</p> <p>Equation of the line: \( y = \sqrt{3}x + (2 - \sqrt{3}) \)</p> <p>Third question: Find the domain and range of the function \( f(x) = \frac{1}{\sqrt{9 - x^2}} \).</p> <p>The domain is where the function is defined and the denominator is not zero.</p> <p>\( 9 - x^2 > 0 \)</p> <p>\( -3 < x < 3 \), so the domain is \( (-3, 3) \).</p> <p>Since the function is the reciprocal of a square root, its range is all positive real numbers, \( (0, \infty) \).</p> <p>Fourth question: Solve the system of equations \( Re(z) = 0, |z| = 2 \).</p> <p>A complex number \( z = x + yi \) satisfies \( Re(z) = 0 \) when \( x = 0 \).</p> <p>Using \( |z| = 2 \), we get \( |0 + yi| = 2 \), which means \( \sqrt{0^2 + y^2} = 2 \).</p> <p>\( y = \pm 2 \), so \( z = 0 \pm 2i \).</p> <p>Fifth question: Write the complex number \( 1 + 7i \) in polar form.</p> <p>Let \( z = 1 + 7i \).</p> <p>Magnitude \( r = |z| = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2} \).</p> <p>Angle \( \theta = \tan^{-1}\left(\frac{7}{1}\right) \), which is in the first quadrant.</p> <p>Polar form: \( z = r(\cos \theta + i \sin \theta) \)</p> <p>\( z = 5\sqrt{2}\left(\cos \left(\tan^{-1} 7\right) + i \sin \left(\tan^{-1} 7\right)\right) \)</p>
Graphing and Analyzing Parabolas
Para la función \( f(x) = 2x^2 - 6x + 4 \), necesitamos determinar la orientación, el eje de simetría, el vértice, los ceros (si los hay), las intercepciones con los ejes, el dominio y el recorrido. Primero, encontramos la fórmula del eje de simetría de una parábola de la forma \( ax^2+bx+c \), que es \( x = -\frac{b}{2a} \). <p>\( x = -\frac{-6}{2(2)} = \frac{6}{4} = \frac{3}{2} \)</p> El eje de simetría es: <p>\( x = \frac{3}{2} \)</p> El vértice de la parábola está en el eje de simetría, así que sustituimos \( x \) por \( \frac{3}{2} \) en \( f(x) \) para encontrar el valor de \( y \) en el vértice. <p>\( f(\frac{3}{2}) = 2(\frac{3}{2})^2 - 6(\frac{3}{2}) + 4 \)</p> <p>\( f(\frac{3}{2}) = 2(\frac{9}{4}) - 9 + 4 \)</p> <p>\( f(\frac{3}{2}) = \frac{18}{4} - \frac{36}{4} + \frac{16}{4} \)</p> <p>\( f(\frac{3}{2}) = -\frac{2}{4} = -\frac{1}{2} \)</p> Entonces, el vértice es: <p>\( V(\frac{3}{2}, -\frac{1}{2}) \)</p> Dado que el coeficiente de \( x^2 \) es positivo, la parábola se abre hacia arriba: <p>Orientación: Hacia arriba</p> Para encontrar los ceros, debemos resolver \( 2x^2 - 6x + 4 = 0 \). Podemos utilizar la fórmula cuadrática o factorizar si es posible. La fórmula cuadrática es \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). <p>\( x = \frac{6 \pm \sqrt{(-6)^2 - 4(2)(4)}}{2(2)} \)</p> <p>\( x = \frac{6 \pm \sqrt{36 - 32}}{4} \)</p> <p>\( x = \frac{6 \pm \sqrt{4}}{4} \)</p> <p>\( x = \frac{6 \pm 2}{4} \)</p> Esto nos da dos ceros: <p>\( x = \frac{8}{4} = 2 \) y \( x = \frac{4}{4} = 1 \)</p> Los ceros son: <p>\( x = 1 \) y \( x = 2 \)</p> El dominio de cualquier parábola es todos los números reales: <p>Dominio: \( (-\infty, \infty) \)</p> El recorrido, dado que la parábola se abre hacia arriba y el vértice es el punto más bajo, será de \( y \) en el vértice (mínimo \( y \)) hasta infinito. <p>Recorrido: \( [-\frac{1}{2}, \infty) \)</p> Las intercepciones con los ejes \( x \) son los ceros: <p>\( (1,0) \) y \( (2,0) \)</p> Para el intercepto con el eje \( y \), evaluamos \( f(0) \): <p>\( f(0) = 2(0)^2 - 6(0) + 4 = 4 \)</p> <p>Intercepto en \( y \): \( (0,4) \)</p>
Determining the Domain and Range from a Graph
<p>The domain of a function is the set of all possible input values (x-values) for which the function is defined, and the range is the set of all possible output values (y-values).</p> <p>Looking at the provided graph, it appears to be a straight line without any breaks or holes, which indicates that the line extends infinitely in both the positive and negative directions along the x-axis.</p> <p>This means the domain of the function is all real numbers.</p> <p>$$ Domain: (-\infty, \infty) $$</p> <p>Similarly, the line extends infinitely in both the positive and negative directions along the y-axis, which means the range of the function is also all real numbers.</p> <p>$$ Range: (-\infty, \infty) $$</p>
Analysis of a Function Using Its Graph
<p>The domain of \( f \): All real numbers \( x \) such that \( -4 \leq x \leq 5 \).</p> <p>The range of \( f \): All real numbers \( y \) such that \( -3 \leq y \leq 5 \).</p> <p>The zeros of \( f \): \( x = -2, 1 \).</p> <p>\( f(-3.5) \): Cannot be determined exactly from the graph; not provided.</p> <p>The intervals on which \( f \) is increasing: \( (-4, -3) \cup (1, 5) \).</p> <p>The intervals on which \( f \) is decreasing: \( (-3, 1) \).</p> <p>The values for which \( f(x) < 0 \): \( x \) in intervals \( (-2, 1) \).</p> <p>Any relative maxima or minima: Relative maximum at \( x = -3 \), relative minimum at \( x = 1 \).</p> <p>The value(s) of \( x \) for which \( f(x) = 3 \): Approximately \( x = 4.5 \).</p> <p>Is \( f(0) \) positive or negative? Negative, since \( f(0) \) is below the \( x \)-axis.</p>
Graph Analysis of a Function
<p>\textbf{(a) The domain of } f:</p> <p>[\text{All real numbers}] \text{, since the graph extends infinitely in the x-direction.}</p> <p>\textbf{(b) The range of } f:</p> <p>[-3, \infty) \text{, because the highest y-value the graph reaches is infinite and the lowest is } -3.</p> <p>\textbf{(c) The zeros of } f:</p> <p> \{ -4, 2 \} \text{, the x-values where the graph intersects the x-axis.}</p> <p>\textbf{(d) } f(-3.5):</p> <p> \text{As } x = -3.5, \text{ f(x) is about } 2.5 \text{, reading from the graph.}</p> <p>\textbf{(e) The intervals on which } f \text{ is increasing:}</p> <p>(-\infty, -4) \cup (2, \infty) \text{, the intervals on the x-axis where the graph goes upwards as x increases.}</p> <p>\textbf{(f) The intervals on which } f \text{ is decreasing:}</p> <p>(-4, 2) \text{, the interval on the x-axis where the graph goes downwards as x increases.}</p> <p>\textbf{(g) The values for which } f(x) \leq 0:</p> <p>[-4, 2] \text{, these are the x-values where the graph is at or below the x-axis.}</p> <p>\textbf{(h) Any relative maxima or minima:}</p> <p>\text{Relative maximum at } (2, 3) \text{, relative minimum at } (-4, -3) \text{ based on the graph's high and low points respectively.}</p> <p>\textbf{(i) The value(s) of } x \text{ for which } f(x) = -3:</p> <p>\{-4, 0\} \text{, the x-values where the graph touches the horizontal line } y = -3.</p> <p>\textbf{(j) Is } f(0) \text{ positive or negative?}</p> <p>\text{Negative, since the point } (0, f(0)) \text{ lies below the x-axis where } f(0) \text{ is around } -3.</p>
Analyzing Functions: Domain, Range, and Functionality
The task is to determine the domain, range, and whether each graph represents a function for each of the six examples. A function is defined as a relation where every input (usually x) has exactly one output (usually y). A common test for functions is the vertical line test. If a vertical line intersects a graph more than once, then the graph does not represent a function. a) For the first graph, which looks like a parabola facing upwards: - The domain is all real numbers since the parabola continues infinitely in both left and right directions. (Domain: ℝ or (-∞, ∞)) - The range is all real numbers greater than or equal to the minimum value of the parabola. Since the vertex is at the x-axis, the minimum y-value is 0. (Range: [0, ∞)) - The graph passes the vertical line test, hence it is a function. b) For the second graph, which is a semicircle: - The domain is the length of the semicircle's base which ranges from -5 to 5. (Domain: [-5, 5]) - The range is the height of the semicircle which ranges from 0 to 5 or the radius of the semicircle. (Range: [0, 5]) - The graph does not pass the vertical line test (any vertical line between x = -5 and x = 5 would intersect the graph twice), hence it is not a function. c) For the third graph, which appears to be a cubic polynomial: - The domain is all real numbers as the graph continues infinitely in both the left and right directions. (Domain: ℝ or (-∞, ∞)) - The range is also all real numbers, as the graph goes infinitely in the upward and downward directions. (Range: ℝ or (-∞, ∞)) - The graph passes the vertical line test, hence it is a function. d) For the fourth graph, which looks like an absolute value function reflected over the x-axis: - The domain is all real numbers as the graph continues infinitely both to the left and right. (Domain: ℝ or (-∞, ∞)) - The range includes all real numbers less than or equal to the maximum value at the x-axis. Since the vertex is at the x-axis, the maximum y-value is 0. (Range: (-∞, 0]) - The graph passes the vertical line test, hence it is a function. e) For the fifth graph, which is a straight line with a negative slope: - The domain is all real numbers, as the line extends infinitely in both left and right directions. (Domain: ℝ or (-∞, ∞)) - The range is also all real numbers, as the line extends infinitely in both the upward and downward directions. (Range: ℝ or (-∞, ∞)) - The graph passes the vertical line test, hence it is a function. f) For the sixth graph, which looks like an arrow going downward: - The domain is all real x-values where the graph is defined. Since the graph is an arrow pointing downwards and stops, the domain is not clearly defined from the image, but it might stop at x = 6 as it appears to. A reasonable assumption would be that domain is all real numbers less than or equal to 6. (Domain: (-∞, 6] or another interval depending on where the graph actually ends.) - The range is the set of y-values that the arrow points to. In this case, it appears to point infinitely downwards, so the range would be all real numbers. (Range: ℝ or (-∞, ∞)) - The graph does not pass the vertical line test since, at x = 6, it appears that multiple y-values are associated with this single x-value. However, since the arrow might indicate a single limit point, without more context or clarity about the graph's intention, it's ambiguous whether this is a function. If the arrow indicates a single point at x = 6, it could still be considered a function. But if the arrow indicates an interval, then it's not a function.
Understanding Function f(x)
The image shows a function f(x) which is defined as: \[ f(x) = \frac{x+1}{x-1} \] In order to help you solve a question involving this function, you'd need to provide me with the actual question. What do you need to do with this function? Are you looking to find the domain, the range, solve for a certain value of x, or something else? Please provide more details.
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