Answer :
Let's go through the problem step-by-step and find the answer to each part.
#### (a) Find the domain of the function.
The domain of a function is the set of all possible input values (x-values) for which the function is defined.
For the given piecewise function:
[tex]\[ f(x)=\left\{\begin{array}{ll} 4+2 x & \text { if } x<0 \\ x^2 & \text { if } x \geq 0 \end{array}\right. \][/tex]
Both pieces of the function are defined for all real numbers. There are no restrictions on [tex]\( x \)[/tex]:
- The first piece, [tex]\( 4 + 2x \)[/tex], is defined for [tex]\( x < 0 \)[/tex].
- The second piece, [tex]\( x^2 \)[/tex], is defined for [tex]\( x \geq 0 \)[/tex].
Combining these, we see that the function is defined for all real numbers. Therefore, the domain of the function [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
This is the solution for part (a).
#### (b) Locate any intercepts.
Intercepts are the points where the function crosses the x-axis and the y-axis.
X-Intercepts:
To find the x-intercepts, we need to solve [tex]\( f(x) = 0 \)[/tex].
1. For [tex]\( x < 0 \)[/tex], we solve:
[tex]\[ 4 + 2x = 0 \][/tex]
[tex]\[ 2x = -4 \][/tex]
[tex]\[ x = -2 \][/tex]
So, one x-intercept is [tex]\((-2, 0)\)[/tex].
2. For [tex]\( x \geq 0 \)[/tex], we solve:
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
So, another x-intercept is [tex]\((0, 0)\)[/tex].
Therefore, the x-intercepts are [tex]\((-2, 0)\)[/tex] and [tex]\((0, 0)\)[/tex].
Y-Intercept:
The y-intercept is found by evaluating [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = 0^2 = 0 \][/tex]
So, the y-intercept is [tex]\((0, 0)\)[/tex].
Containing all this information, the intercepts are:
[tex]\[ \text{x-intercepts: } [(-2, 0), (0, 0)] \][/tex]
[tex]\[ \text{y-intercept: } (0, 0) \][/tex]
#### (c) Graph the function.
To graph the function, notice that it is piecewise:
- For [tex]\( x < 0 \)[/tex], the function [tex]\( f(x) = 4 + 2x \)[/tex] is a linear function with a slope of 2 and a y-intercept at [tex]\( (0, 4) \)[/tex]. This part of the graph will be a straight line that decreases as [tex]\( x \)[/tex] becomes more negative.
- For [tex]\( x \geq 0 \)[/tex], the function [tex]\( f(x) = x^2 \)[/tex] is a quadratic (parabolic) function that opens upward with its vertex at the origin [tex]\( (0, 0) \)[/tex].
You can sketch these together ensuring there is continuity at [tex]\( x = 0 \)[/tex].
#### (d) Based on the graph, find the range.
The range of a function is the set of all possible output values (y-values).
- For [tex]\( x < 0 \)[/tex], the function [tex]\( f(x) = 4 + 2x \)[/tex] decreases without bound as [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]). Therefore, [tex]\( f(x) \)[/tex] takes all values less than 4.
- For [tex]\( x \geq 0 \)[/tex], the function [tex]\( f(x) = x^2 \)[/tex] increases without bound as [tex]\( x \)[/tex] approaches positive infinity ([tex]\( \infty \)[/tex]). Therefore, [tex]\( f(x) \)[/tex] takes all values greater than or equal to 0.
Since the two pieces join at [tex]\( (0, 0) \)[/tex], combining these observations:
- The lowest value [tex]\( f(x) \)[/tex] can take is [tex]\( -\infty \)[/tex], and the highest unbounded value it can take is [tex]\( \infty \)[/tex].
Thus, the range of the function [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
This is the detailed, step-by-step solution to each part of the given question.
#### (a) Find the domain of the function.
The domain of a function is the set of all possible input values (x-values) for which the function is defined.
For the given piecewise function:
[tex]\[ f(x)=\left\{\begin{array}{ll} 4+2 x & \text { if } x<0 \\ x^2 & \text { if } x \geq 0 \end{array}\right. \][/tex]
Both pieces of the function are defined for all real numbers. There are no restrictions on [tex]\( x \)[/tex]:
- The first piece, [tex]\( 4 + 2x \)[/tex], is defined for [tex]\( x < 0 \)[/tex].
- The second piece, [tex]\( x^2 \)[/tex], is defined for [tex]\( x \geq 0 \)[/tex].
Combining these, we see that the function is defined for all real numbers. Therefore, the domain of the function [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
This is the solution for part (a).
#### (b) Locate any intercepts.
Intercepts are the points where the function crosses the x-axis and the y-axis.
X-Intercepts:
To find the x-intercepts, we need to solve [tex]\( f(x) = 0 \)[/tex].
1. For [tex]\( x < 0 \)[/tex], we solve:
[tex]\[ 4 + 2x = 0 \][/tex]
[tex]\[ 2x = -4 \][/tex]
[tex]\[ x = -2 \][/tex]
So, one x-intercept is [tex]\((-2, 0)\)[/tex].
2. For [tex]\( x \geq 0 \)[/tex], we solve:
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
So, another x-intercept is [tex]\((0, 0)\)[/tex].
Therefore, the x-intercepts are [tex]\((-2, 0)\)[/tex] and [tex]\((0, 0)\)[/tex].
Y-Intercept:
The y-intercept is found by evaluating [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = 0^2 = 0 \][/tex]
So, the y-intercept is [tex]\((0, 0)\)[/tex].
Containing all this information, the intercepts are:
[tex]\[ \text{x-intercepts: } [(-2, 0), (0, 0)] \][/tex]
[tex]\[ \text{y-intercept: } (0, 0) \][/tex]
#### (c) Graph the function.
To graph the function, notice that it is piecewise:
- For [tex]\( x < 0 \)[/tex], the function [tex]\( f(x) = 4 + 2x \)[/tex] is a linear function with a slope of 2 and a y-intercept at [tex]\( (0, 4) \)[/tex]. This part of the graph will be a straight line that decreases as [tex]\( x \)[/tex] becomes more negative.
- For [tex]\( x \geq 0 \)[/tex], the function [tex]\( f(x) = x^2 \)[/tex] is a quadratic (parabolic) function that opens upward with its vertex at the origin [tex]\( (0, 0) \)[/tex].
You can sketch these together ensuring there is continuity at [tex]\( x = 0 \)[/tex].
#### (d) Based on the graph, find the range.
The range of a function is the set of all possible output values (y-values).
- For [tex]\( x < 0 \)[/tex], the function [tex]\( f(x) = 4 + 2x \)[/tex] decreases without bound as [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]). Therefore, [tex]\( f(x) \)[/tex] takes all values less than 4.
- For [tex]\( x \geq 0 \)[/tex], the function [tex]\( f(x) = x^2 \)[/tex] increases without bound as [tex]\( x \)[/tex] approaches positive infinity ([tex]\( \infty \)[/tex]). Therefore, [tex]\( f(x) \)[/tex] takes all values greater than or equal to 0.
Since the two pieces join at [tex]\( (0, 0) \)[/tex], combining these observations:
- The lowest value [tex]\( f(x) \)[/tex] can take is [tex]\( -\infty \)[/tex], and the highest unbounded value it can take is [tex]\( \infty \)[/tex].
Thus, the range of the function [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
This is the detailed, step-by-step solution to each part of the given question.
