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Sagot :
Certainly! Let's go through the steps to graph the function [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex] and identify its domain and range.
### Step 1: Understand the Function
The given function is [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex]. This is a quadratic function, and its general shape is a parabola.
### Step 2: Determine the Domain
The domain of a function refers to all the possible values of [tex]\( x \)[/tex] for which the function is defined. For the quadratic function [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex], there are no restrictions on [tex]\( x \)[/tex]. Thus, the domain is all real numbers:
[tex]\[ \text{Domain} = (-\infty, \infty) \][/tex]
### Step 3: Determine the Range
The range of a function refers to all the possible values of [tex]\( f(x) \)[/tex]. Since [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex] is a parabola that opens upwards, the vertex of the parabola (which is at the origin, [tex]\( (0, 0) \)[/tex]) is the minimum point. The function's value will always be non-negative because a square of any real number is non-negative.
Thus, the range is all non-negative real numbers:
[tex]\[ \text{Range} = [0, \infty) \][/tex]
### Step 4: Plot the Function
To plot the function:
1. Choose a set of [tex]\( x \)[/tex]-values. A suitable range might be from [tex]\(-10\)[/tex] to [tex]\(10\)[/tex], since we're aiming to get a reasonably perceptible shape for the parabola.
2. Calculate the corresponding [tex]\( y \)[/tex]-values (i.e., [tex]\( f(x) \)[/tex]-values).
For example:
- When [tex]\( x = -2 \)[/tex], [tex]\( f(x) = \frac{1}{2}(-2)^2 = 2 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = \frac{1}{2}(0)^2 = 0 \)[/tex].
- When [tex]\( x = 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}(2)^2 = 2 \)[/tex].
These points will help us sketch the graph.
### Step 5: Draw the Graph
Here is a rough sketch of the function [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex]:
[tex]\[ \begin{array}{c|c} x & f(x) = \frac{1}{2}x^2 \\ \hline -3 & 4.5 \\ -2 & 2 \\ -1 & 0.5 \\ 0 & 0 \\ 1 & 0.5 \\ 2 & 2 \\ 3 & 4.5 \\ \end{array} \][/tex]
Plot these points on a coordinate plane:
- Point at [tex]\( (-3, 4.5) \)[/tex]
- Point at [tex]\( (-2, 2) \)[/tex]
- Point at [tex]\( (-1, 0.5) \)[/tex]
- Point at [tex]\( (0, 0) \)[/tex]
- Point at [tex]\( (1, 0.5) \)[/tex]
- Point at [tex]\( (2, 2) \)[/tex]
- Point at [tex]\( (3, 4.5) \)[/tex]
Now, connect these points with a smooth curve (parabola). The graph will show that as [tex]\( x \)[/tex] moves away from 0 in positive or negative direction, [tex]\( f(x) \)[/tex] increases.
### Conclusion
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
- Range: [tex]\( [0, \infty) \)[/tex]
The graph of [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex] is an upward-opening parabola with its vertex at the origin (0,0).
### Step 1: Understand the Function
The given function is [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex]. This is a quadratic function, and its general shape is a parabola.
### Step 2: Determine the Domain
The domain of a function refers to all the possible values of [tex]\( x \)[/tex] for which the function is defined. For the quadratic function [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex], there are no restrictions on [tex]\( x \)[/tex]. Thus, the domain is all real numbers:
[tex]\[ \text{Domain} = (-\infty, \infty) \][/tex]
### Step 3: Determine the Range
The range of a function refers to all the possible values of [tex]\( f(x) \)[/tex]. Since [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex] is a parabola that opens upwards, the vertex of the parabola (which is at the origin, [tex]\( (0, 0) \)[/tex]) is the minimum point. The function's value will always be non-negative because a square of any real number is non-negative.
Thus, the range is all non-negative real numbers:
[tex]\[ \text{Range} = [0, \infty) \][/tex]
### Step 4: Plot the Function
To plot the function:
1. Choose a set of [tex]\( x \)[/tex]-values. A suitable range might be from [tex]\(-10\)[/tex] to [tex]\(10\)[/tex], since we're aiming to get a reasonably perceptible shape for the parabola.
2. Calculate the corresponding [tex]\( y \)[/tex]-values (i.e., [tex]\( f(x) \)[/tex]-values).
For example:
- When [tex]\( x = -2 \)[/tex], [tex]\( f(x) = \frac{1}{2}(-2)^2 = 2 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = \frac{1}{2}(0)^2 = 0 \)[/tex].
- When [tex]\( x = 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}(2)^2 = 2 \)[/tex].
These points will help us sketch the graph.
### Step 5: Draw the Graph
Here is a rough sketch of the function [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex]:
[tex]\[ \begin{array}{c|c} x & f(x) = \frac{1}{2}x^2 \\ \hline -3 & 4.5 \\ -2 & 2 \\ -1 & 0.5 \\ 0 & 0 \\ 1 & 0.5 \\ 2 & 2 \\ 3 & 4.5 \\ \end{array} \][/tex]
Plot these points on a coordinate plane:
- Point at [tex]\( (-3, 4.5) \)[/tex]
- Point at [tex]\( (-2, 2) \)[/tex]
- Point at [tex]\( (-1, 0.5) \)[/tex]
- Point at [tex]\( (0, 0) \)[/tex]
- Point at [tex]\( (1, 0.5) \)[/tex]
- Point at [tex]\( (2, 2) \)[/tex]
- Point at [tex]\( (3, 4.5) \)[/tex]
Now, connect these points with a smooth curve (parabola). The graph will show that as [tex]\( x \)[/tex] moves away from 0 in positive or negative direction, [tex]\( f(x) \)[/tex] increases.
### Conclusion
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
- Range: [tex]\( [0, \infty) \)[/tex]
The graph of [tex]\( f(x) = \frac{1}{2} x^2 \)[/tex] is an upward-opening parabola with its vertex at the origin (0,0).
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