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Chapter 0: Problem 32

Find the domain and sketch the graph of the function. What is its range? $$ f(x)=\frac{1}{2} x^{2}+1 $$

Short Answer

Expert verified
The domain of the function \(f(x) = \frac{1}{2}x^2 + 1\) is \(-\infty < x < \infty\). The graph of the function is a parabola opening upward with its vertex at the point (0,1). The range of the function is \(1 \leq y < \infty\).

Step by step solution

01

Determine the domain of the function

To find the domain, we need to find the set of all x-values for which the function is defined. Since the function is a quadratic function, it is defined for all real values of x. Therefore, the domain of the function is: Domain: \(-\infty < x < \infty\)
02

Analyze the properties of the function

As mentioned earlier, the function is a quadratic function. Since the leading coefficient is positive (\(\frac{1}{2}\)), the parabola will open upward. The vertex of the parabola is the point at which the function achieves its minimum value. The general form of a parabola is given by \(f(x) = ax^2 + bx + c\), and the x-coordinate of the vertex can be found using the formula \(x_V = -\frac{b}{2a}\). In our case, we have \(a=\frac{1}{2}\), \(b=0\), and \(c=1\).
03

Calculate the vertex of the parabola

Using the formula mentioned above (\(x_V = -\frac{b}{2a}\)), we have: \(x_V = -\frac{0}{2\cdot(1/2)} = 0\) Now, we need to find the corresponding y-coordinate of the vertex. This can be done by plugging the x-coordinate into the function: \(y_V = f(x_V) = \frac{1}{2}\cdot(0)^2 + 1 = 1\) So the vertex of the parabola is at point (0,1).
04

Sketch the graph of the function

From our analysis, we know that the graph of the function is a parabola that opens upward with its vertex at the point (0,1). To sketch the graph, we can plot some more points by choosing different x-values and finding the corresponding y-values: \(f(-1) = \frac{1}{2}\cdot(-1)^2 + 1 = \frac{3}{2}\) \(f(-2) = \frac{1}{2}\cdot(-2)^2 + 1 = 3\) \(f(1) = \frac{1}{2}\cdot(1)^2 + 1 = \frac{3}{2}\) \(f(2) = \frac{1}{2}\cdot(2)^2 + 1 = 3\) Plot these points and the vertex on the coordinate plane, then draw a smooth curve to connect the points.
05

Determine the range of the function

Since the parabola opens upward and has its vertex at (0,1), the function will have a minimum value of 1 at the vertex. The function will increase in both directions from the vertex. Hence, the range of the function will be all values greater than or equal to 1: Range: \(1 \leq y < \infty\) So, the domain of the function is \(-\infty < x < \infty\), its range is \(1 \leq y < \infty\), and the graph of the function is a parabola opening upward with its vertex at the point (0,1).

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