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Sagot :
To determine which of the given functions have a range of [tex]\(\{y \in \mathbb{R} \mid -\infty
1. [tex]\(f(x) = -4x + 11\)[/tex]
This is a linear function with a slope of [tex]\(-4\)[/tex] and a y-intercept of [tex]\(11\)[/tex]. Linear functions of the form [tex]\(ax + b\)[/tex], where [tex]\(a \neq 0\)[/tex], have a range of all real numbers. Thus, this function's range is [tex]\( \{y \in \mathbb{R} \mid -\infty
2. [tex]\(f(x) = -(x+1)^2 - 4\)[/tex]
This is a quadratic function in the form of [tex]\(y = ax^2 + bx + c\)[/tex] where [tex]\(a < 0\)[/tex] (specifically, [tex]\(a=-1\)[/tex]). Because it is a downward-opening parabola, it has a maximum value at the vertex and extends downward to [tex]\(-\infty\)[/tex]. The vertex is at [tex]\(x=-1\)[/tex], and the maximum value is:
[tex]\[f(-1) = -((-1+1)^2) - 4 = -0 - 4 = -4\][/tex]
Thus, the range of the function is [tex]\( \{y \leq -4 \mid y \in \mathbb{R}\} \)[/tex]. This function does not cover all real numbers.
3. [tex]\(f(x) = \frac{2}{3}x - 8\)[/tex]
Similar to the first function, this is another linear function. It has a slope of [tex]\(\frac{2}{3}\)[/tex] and a y-intercept of [tex]\(-8\)[/tex]. As mentioned before, linear functions with a non-zero slope have a range of all real numbers, [tex]\( \{y \in \mathbb{R} \mid -\infty
4. [tex]\(f(x) = x^2 + 7x - 9\)[/tex]
This is a quadratic function opening upwards (since the coefficient of [tex]\(x^2\)[/tex] is positive). Quadratics opening upwards achieve their minimum value at the vertex but extend upwards to [tex]\(\infty\)[/tex]. This means the range is from the vertex's minimum value to [tex]\(\infty\)[/tex]. The minimum value is at:
[tex]\[x = -\frac{b}{2a} = -\frac{7}{2 \cdot 1} = -\frac{7}{2}\][/tex]
Calculating the function value at this point:
[tex]\[f\left(-\frac{7}{2}\right) = \left(-\frac{7}{2}\right)^2 + 7\left(-\frac{7}{2}\right) - 9 = \frac{49}{4} - \frac{49}{2} - 9 = \frac{49}{4} - \frac{98}{4} - \frac{36}{4} = \frac{49 - 98 - 36}{4} = \frac{-85}{4}\][/tex]
Thus, the range is [tex]\(\left\{ y \in \mathbb{R} \mid y \geq \frac{-85}{4} \right\}\)[/tex]. Therefore, it does not cover all real numbers.
5. [tex]\(f(x) = 2^{x+3}\)[/tex]
This is an exponential function. Exponential functions of the form [tex]\(a^x\)[/tex] where [tex]\(a > 1\)[/tex] have a range of [tex]\( (0, \infty) \)[/tex]. This means the function only takes positive values, no negatives, and does not extend to all real numbers.
In conclusion, the functions that have a range of [tex]\(\{y \in \mathbb{R} \mid -\infty < y < \infty\}\)[/tex] are:
[tex]\[ f(x) = -4x + 11 \][/tex]
[tex]\[ f(x) = \frac{2}{3}x - 8 \][/tex]
1. [tex]\(f(x) = -4x + 11\)[/tex]
This is a linear function with a slope of [tex]\(-4\)[/tex] and a y-intercept of [tex]\(11\)[/tex]. Linear functions of the form [tex]\(ax + b\)[/tex], where [tex]\(a \neq 0\)[/tex], have a range of all real numbers. Thus, this function's range is [tex]\( \{y \in \mathbb{R} \mid -\infty
2. [tex]\(f(x) = -(x+1)^2 - 4\)[/tex]
This is a quadratic function in the form of [tex]\(y = ax^2 + bx + c\)[/tex] where [tex]\(a < 0\)[/tex] (specifically, [tex]\(a=-1\)[/tex]). Because it is a downward-opening parabola, it has a maximum value at the vertex and extends downward to [tex]\(-\infty\)[/tex]. The vertex is at [tex]\(x=-1\)[/tex], and the maximum value is:
[tex]\[f(-1) = -((-1+1)^2) - 4 = -0 - 4 = -4\][/tex]
Thus, the range of the function is [tex]\( \{y \leq -4 \mid y \in \mathbb{R}\} \)[/tex]. This function does not cover all real numbers.
3. [tex]\(f(x) = \frac{2}{3}x - 8\)[/tex]
Similar to the first function, this is another linear function. It has a slope of [tex]\(\frac{2}{3}\)[/tex] and a y-intercept of [tex]\(-8\)[/tex]. As mentioned before, linear functions with a non-zero slope have a range of all real numbers, [tex]\( \{y \in \mathbb{R} \mid -\infty
4. [tex]\(f(x) = x^2 + 7x - 9\)[/tex]
This is a quadratic function opening upwards (since the coefficient of [tex]\(x^2\)[/tex] is positive). Quadratics opening upwards achieve their minimum value at the vertex but extend upwards to [tex]\(\infty\)[/tex]. This means the range is from the vertex's minimum value to [tex]\(\infty\)[/tex]. The minimum value is at:
[tex]\[x = -\frac{b}{2a} = -\frac{7}{2 \cdot 1} = -\frac{7}{2}\][/tex]
Calculating the function value at this point:
[tex]\[f\left(-\frac{7}{2}\right) = \left(-\frac{7}{2}\right)^2 + 7\left(-\frac{7}{2}\right) - 9 = \frac{49}{4} - \frac{49}{2} - 9 = \frac{49}{4} - \frac{98}{4} - \frac{36}{4} = \frac{49 - 98 - 36}{4} = \frac{-85}{4}\][/tex]
Thus, the range is [tex]\(\left\{ y \in \mathbb{R} \mid y \geq \frac{-85}{4} \right\}\)[/tex]. Therefore, it does not cover all real numbers.
5. [tex]\(f(x) = 2^{x+3}\)[/tex]
This is an exponential function. Exponential functions of the form [tex]\(a^x\)[/tex] where [tex]\(a > 1\)[/tex] have a range of [tex]\( (0, \infty) \)[/tex]. This means the function only takes positive values, no negatives, and does not extend to all real numbers.
In conclusion, the functions that have a range of [tex]\(\{y \in \mathbb{R} \mid -\infty < y < \infty\}\)[/tex] are:
[tex]\[ f(x) = -4x + 11 \][/tex]
[tex]\[ f(x) = \frac{2}{3}x - 8 \][/tex]
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