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Sure! Let's analyze and solve this problem step-by-step.
### a. Graph the Function
We are given a piecewise function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \begin{cases} \frac{1}{4} x^2 & \text{if } x < 4 \\ 4x - 10 & \text{if } x \geq 4 \end{cases} \][/tex]
To graph the function, let's consider the two parts of the piecewise function separately:
1. For [tex]\( x < 4 \)[/tex]:
The function is [tex]\( f(x) = \frac{1}{4} x^2 \)[/tex], which is a quadratic function opening upwards.
2. For [tex]\( x \geq 4 \)[/tex]:
The function is [tex]\( f(x) = 4x - 10 \)[/tex], which is a linear function with a slope of 4 and a y-intercept of -10.
Let's find the point at which these two parts meet:
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = \frac{1}{4} 4^2 = 4 \][/tex]
[tex]\[ f(4) = 4 4 - 10 = 6 \][/tex]
Since we have the boundary case [tex]\( x = 4 \)[/tex], the correct value at [tex]\( x = 4 \)[/tex] should be taken from the second part of the function, which serves [tex]\( x \geq 4 \)[/tex]. Hence, at [tex]\( x = 4 \)[/tex],
[tex]\[ f(4) = 6 \][/tex]
Now, we can plot the two parts:
- The quadratic function [tex]\( \frac{1}{4} x^2 \)[/tex] for [tex]\( x < 4 \)[/tex]
- The linear function [tex]\( 4x - 10 \)[/tex] for [tex]\( x \geq 4 \)[/tex]
### b. Determine the Function's Range
To determine the range of this piecewise function using the graph, we need to understand the value [tex]\( f(x) \)[/tex] takes for all possible [tex]\( x \)[/tex].
1. For [tex]\( x < 4 \)[/tex]:
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]), [tex]\( f(x) \)[/tex] approaches infinity ([tex]\( +\infty \)[/tex]) because of the [tex]\( \frac{1}{4} x^2 \)[/tex] quadratic term.
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = \frac{1}{4} * 0^2 = 0 \)[/tex].
2. For [tex]\( x \geq 4 \)[/tex]:
- The value immediately at [tex]\( x = 4 \)[/tex] is calculated from the second piece, so [tex]\( f(4) = 6 \)[/tex].
- As [tex]\( x \)[/tex] increases beyond 4, since [tex]\( 4x - 10 \)[/tex] is a linear function with a positive slope, [tex]\( f(x) \)[/tex] will continue increasing without bound.
From these observations, we can see:
- The lowest value of the function occurs at [tex]\( x = 0 \)[/tex] with [tex]\( f(0) = 0 \)[/tex].
- The function increases in both directions from zero to [tex]\( +\infty \)[/tex].
Therefore, the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ [0, \infty) \][/tex]
Here is a summary of the findings:
- The function value starts at 0 and increases to positive infinity.
- The range of [tex]\( f(x) \)[/tex]: [tex]\( [0, \infty) \)[/tex].
### a. Graph the Function
We are given a piecewise function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \begin{cases} \frac{1}{4} x^2 & \text{if } x < 4 \\ 4x - 10 & \text{if } x \geq 4 \end{cases} \][/tex]
To graph the function, let's consider the two parts of the piecewise function separately:
1. For [tex]\( x < 4 \)[/tex]:
The function is [tex]\( f(x) = \frac{1}{4} x^2 \)[/tex], which is a quadratic function opening upwards.
2. For [tex]\( x \geq 4 \)[/tex]:
The function is [tex]\( f(x) = 4x - 10 \)[/tex], which is a linear function with a slope of 4 and a y-intercept of -10.
Let's find the point at which these two parts meet:
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = \frac{1}{4} 4^2 = 4 \][/tex]
[tex]\[ f(4) = 4 4 - 10 = 6 \][/tex]
Since we have the boundary case [tex]\( x = 4 \)[/tex], the correct value at [tex]\( x = 4 \)[/tex] should be taken from the second part of the function, which serves [tex]\( x \geq 4 \)[/tex]. Hence, at [tex]\( x = 4 \)[/tex],
[tex]\[ f(4) = 6 \][/tex]
Now, we can plot the two parts:
- The quadratic function [tex]\( \frac{1}{4} x^2 \)[/tex] for [tex]\( x < 4 \)[/tex]
- The linear function [tex]\( 4x - 10 \)[/tex] for [tex]\( x \geq 4 \)[/tex]
### b. Determine the Function's Range
To determine the range of this piecewise function using the graph, we need to understand the value [tex]\( f(x) \)[/tex] takes for all possible [tex]\( x \)[/tex].
1. For [tex]\( x < 4 \)[/tex]:
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]), [tex]\( f(x) \)[/tex] approaches infinity ([tex]\( +\infty \)[/tex]) because of the [tex]\( \frac{1}{4} x^2 \)[/tex] quadratic term.
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = \frac{1}{4} * 0^2 = 0 \)[/tex].
2. For [tex]\( x \geq 4 \)[/tex]:
- The value immediately at [tex]\( x = 4 \)[/tex] is calculated from the second piece, so [tex]\( f(4) = 6 \)[/tex].
- As [tex]\( x \)[/tex] increases beyond 4, since [tex]\( 4x - 10 \)[/tex] is a linear function with a positive slope, [tex]\( f(x) \)[/tex] will continue increasing without bound.
From these observations, we can see:
- The lowest value of the function occurs at [tex]\( x = 0 \)[/tex] with [tex]\( f(0) = 0 \)[/tex].
- The function increases in both directions from zero to [tex]\( +\infty \)[/tex].
Therefore, the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ [0, \infty) \][/tex]
Here is a summary of the findings:
- The function value starts at 0 and increases to positive infinity.
- The range of [tex]\( f(x) \)[/tex]: [tex]\( [0, \infty) \)[/tex].
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