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Chapter 3: Problem 14

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ g(x)=\frac{1}{2} x-\sqrt{x} $$

Short Answer

Expert verified
The domain of the function \(g(x) = \frac{1}{2}x - \sqrt{x}\) is \(x \geq 0\). The x-intercept is x = 0, and the y-intercept is g(0) = 0. There are no vertical or horizontal asymptotes. The function has a critical point when \(g'(x) = \frac{1}{2} - \frac{1}{2\sqrt{x}} = 0\), which occurs at \(x = 1\). The function is decreasing on the interval \(0 \leq x < 1\) and increasing on the interval \(1 < x \leq \infty\). There is a local minimum point at \(x = 1\), and the graph looks like a parabola that opens to the right.

Step by step solution

01

1. Finding the domain

We determine the values of x for which the function is defined. There are no restrictions on x for the linear part of the function (1/2)x, but the square root function is defined only for non-negative values for x. Therefore, the domain of g(x) is \(x \geq 0\).
02

2. Finding x and y-intercepts

To find the x-intercept, let g(x) = 0 and solve for x: \(0 = \frac{1}{2}x - \sqrt{x}\) To find the y-intercept, let x = 0 and solve for g(x): \(g(0) = y\)
03

3. Finding vertical and horizontal asymptotes

There are no vertical asymptotes for this function because there are no restrictions on the domain, and there are no terms in the function that would give us a vertical asymptote. There are also no horizontal asymptotes because the function doesn't reach a constant value as \(x \to \infty\).
04

4. Finding intervals of increasing and decreasing

To find the intervals of increasing and decreasing, we need to find the first derivative and its critical points: \(g'(x) = \frac{1}{2} - \frac{1}{2\sqrt{x}}\) Now, we find the critical points by setting g'(x) to zero: \(0 = \frac{1}{2} - \frac{1}{2\sqrt{x}}\) With the critical points, we can determine the intervals of increasing and decreasing.
05

5. Finding local maximum and minimum points

The local maximum and minimum points can be found by analyzing the critical points and determining whether g'(x) changes its sign from positive to negative (local maximum) or negative to positive (local minimum).
06

6. Sketching the graph

With all the information gathered above: domain, intercepts, asymptotes, intervals of increasing/decreasing, and local maximum/minimum points, we can now sketch the graph of the function g(x) = (1/2)x - 鈭歺 using these characteristics as a guideline. Remember to label important points (intercepts, critical points, minimum/maximum points) and indicate the intervals of increasing and decreasing on your sketch.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain and Range
In function analysis, the domain and range are fundamental concepts that help in understanding the set of possible inputs and outputs. For the function \(g(x) = \frac{1}{2} x - \sqrt{x}\), let's explore its domain and range.
  • Domain: The domain refers to all values of \(x\) for which the function is defined. Since \(\sqrt{x}\) is defined only for non-negative values, the domain of \(g(x)\) is \(x \geq 0\). This means that \(x\) can take any value from zero to positive infinity.
  • Range: The range is the set of all possible output values of \(g(x)\). To find the range, it helps to consider the behavior of the function as \(x\) varies within its domain. As \(x\) increases from zero, \(\frac{1}{2}x\) grows linearly while \(-\sqrt{x}\) becomes more negative. Thus, the range is all real numbers \(y\) which \(g(x)\) will attain by different \(x\).
Understanding the domain and range gives us a clearer picture of where the function exists and behaves, which is crucial for graphing.
Critical Points
Critical points are where a function demonstrates special behavior, like turning points or points of interest. For \(g(x)\), identifying these can elucidate the graph's key features.
  • To find critical points, we compute the first derivative \(g'(x)\).
  • We set \(g'(x) = 0\) to find where the function's rate of change is zero: \(0 = \frac{1}{2} - \frac{1}{2\sqrt{x}}\).
  • Solving this equation gives us the critical points.
These points are essential because they specify where the graph reaches peaks or troughs, allowing us to determine the increasing or decreasing nature of the function in different intervals.
Function Analysis
Function analysis involves breaking down a function to understand it completely. For \(g(x)\), this means exploring different aspects like intercepts, asymptotes, and intervals of behavior.
  • Intercepts: The x-intercept is found by setting \(g(x) = 0\), leading to solving the equation \(\frac{1}{2}x - \sqrt{x} = 0\). The y-intercept is found by evaluating \(g(0)\).
  • Asymptotes: Examining the behavior at infinity, \(g(x)\) does not have vertical or horizontal asymptotes because it doesn鈥檛 approach a single constant value or become undefined.
  • Intervals of Increasing/Decreasing: By assessing \(g'(x)\), we determine where \(g(x)\) rises or falls. The sign of the derivative tells whether the function is increasing or decreasing in those sections.
Through thorough function analysis, we gather a complete picture of the function's graphical behavior.
Derivatives
Derivatives provide a tool for understanding the rate of change of a function. For \(g(x) = \frac{1}{2} x - \sqrt{x}\), the derivative \(g'(x)\) offers insight into its slope and behavior.
  • The first derivative \(g'(x) = \frac{1}{2} - \frac{1}{2\sqrt{x}}\) reveals how \(g(x)\) changes as \(x\) changes.
  • It highlights critical points, where the graph shifts from increasing to decreasing or vice versa, by setting it to zero.
  • Additionally, the derivative informs us about the slope of the tangent to the graph at any given point, which is crucial for sketching.
Understanding derivatives is key to uncovering the detailed behavior of the function, and they form the backbone of calculus-based function analysis.

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