Chapter 15: Problem 356
Determine the extremes of the expression \(\mathrm{f}(\mathrm{x})=\sqrt{\mid \mathrm{x}} \mid\) and sketch the graph.
Short Answer
Expert verified
The function \(f(x) = \sqrt{|x|}\) has no local maxima or minima, and its graph is symmetric about the y-axis. The graph is strictly increasing for x ≥ 0 and strictly decreasing for x < 0. It has a "V" shape, with a vertex at the origin (0, 0) and branches extending into quadrants I and II.
Step by step solution
01
Find the first derivative of f(x)
To find the first derivative, we first rewrite the function f(x) as:
\(f(x) = √(|x|)\)
Now, since f(x) is an even function, we can rewrite it as:
\(f(x) = \left\{
\begin{array}{ll}
√x & for \: x\geq0 \\
√(-x) & for \: x<0 \\
\end{array}
\right.\)
Now we can find the first derivative in each domain:
For x ≥ 0:
\(f'(x) = \frac{1}{2}x^{-\frac{1}{2}}\)
For x < 0:
\(f'(x) = -\frac{1}{2}(-x)^{-\frac{1}{2}}\)
02
Identify critical points and classify them as maxima, minima or saddle points
Since the function consists of two different parts, we have to consider them separately.
For x ≥ 0:
\(f'(x) = \frac{1}{2}x^{-\frac{1}{2}}\)
The derivative is always positive, so the function is strictly increasing in this domain. Therefore, there are no local maxima or minima in x ≥ 0.
For x < 0:
\(f'(x) = -\frac{1}{2}(-x)^{-\frac{1}{2}}\)
The derivative is always negative, so the function is strictly decreasing in this domain. Therefore, there are no local maxima or minima in x < 0.
As the function is strictly increasing and decreasing in each domain respectively, and the continuous at x = 0, we can conclude that there are no local maxima or minima, meaning there are no extremes for this function.
03
Sketch the graph of f(x) = √|x|
Since we have no extremes, we will sketch the graph using the information we have on the function. The graph will consist of two different parts:
1. For x ≥ 0, the function is strictly increasing and is equal to √x. This means that the graph will have a vertical tangent at x = 0 and continue to increase as x increases.
2. For x < 0, the function is strictly decreasing and is equal to √(-x). This means that the graph will have a vertical tangent at x = 0 and continue to decrease as x decreases.
The graph of f(x) = √|x| will be symmetric about the y-axis since it's an even function. The final graph will look like a "V" shape, opening upwards, with the vertex at the origin (0, 0), and branches extending into quadrants I and II.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are at the heart of calculus, serving as a tool to determine the rate at which a function changes at any given point. For a given function, the derivative helps reveal important characteristics about its behavior. In the function \(f(x) = \sqrt{|x|}\), we first need to express it piecewise to apply derivatives correctly. This involves considering two cases for the expression based on the value of \(x\):
- When \(x \geq 0\), the function simplifies to \(\sqrt{x}\).
- When \(x < 0\), the function simplifies to \(\sqrt{-x}\).
Critical Points
Critical points occur where the first derivative of a function is zero or undefined. They are pivotal in determining where a function might have a local maximum, minimum, or saddle point. However, in our function \(f(x) = \sqrt{|x|}\), the derivative is never equal to zero for \(x eq 0\).
It's intrinsic to note that for this particular function, \(x = 0\) is a point of interest. At this point, the derivative isn't zero but the function changes from decreasing to increasing.
It's intrinsic to note that for this particular function, \(x = 0\) is a point of interest. At this point, the derivative isn't zero but the function changes from decreasing to increasing.
- For \(x \geq 0\), the derivative is positive, suggesting the function is increasing.
- For \(x < 0\), the derivative is negative, indicating the function is decreasing.
Graph Sketching
Graph sketching is a valuable skill that enables visualization of mathematical functions. For \(f(x) = \sqrt{|x|}\), sketching the graph involves using all derived information. We divide the task based on the domains:
- For \(x \geq 0\), the function behaves like \(\sqrt{x}\) which is an increasing curve starting from the origin.
- For \(x < 0\), the function behaves like \(\sqrt{-x}\) making it a decreasing curve again originating from the origin.
Even Functions
Even functions have the characteristic property of symmetry about the y-axis. For a function \(f(x)\), this implies that \(f(x) = f(-x)\). Our discussed function \(f(x) = \sqrt{|x|}\) clearly embodies this property, as changing \(x\) to \(-x\) does not alter the outcome of the function.
This symmetrical property is essential for understanding the graph's shape, as it guarantees a mirrored behavior across the y-axis, contributing to its unique V-shape.
This symmetrical property is essential for understanding the graph's shape, as it guarantees a mirrored behavior across the y-axis, contributing to its unique V-shape.
- When sketching, this means that we only need to evaluate one half of the graph and then reflect it over the y-axis to complete it.
- This reflection about the y-axis simplifies analysis and computation in symmetrical functions.