Problem 70
Question
\begin{equation} \begin{array}{l}{\text { a. The function } y=\tan x+3 \cot x \text { has an absolute minimum }} \\ {\text { value on the interval } 0 < x < \pi / 2 . \text { Find it. }} \\ {\text { b. Graph the function and compare what you see with your }} \\ {\text { answer in part (a). }}\end{array} \end{equation}
Step-by-Step Solution
Verified Answer
The absolute minimum value is \( 2\sqrt{3} \).
1Step 1: Understanding the Function
The given function is \( y = \tan x + 3 \cot x \). We need to find the absolute minimum value of this function on the interval \( 0 < x < \frac{\pi}{2} \). Both \( \tan x \) and \( \cot x \) are undefined at \( x = 0 \) and \( x = \frac{\pi}{2} \), but are continuous and differentiable in between.
2Step 2: Compute the Derivative
To find critical points, we need to compute the derivative of the function. The derivative of \( \tan x \) is \( \sec^2 x \), and the derivative of \( 3 \cot x \) is \(-3 \csc^2 x \). Therefore, the derivative of \( y \) is: \[ y' = \sec^2 x - 3 \csc^2 x. \]
3Step 3: Find Critical Points
Set the derivative \( y' \) equal to zero to find critical points:\[ \sec^2 x - 3 \csc^2 x = 0 \]\[ \sec^2 x = 3 \csc^2 x \]\[ \frac{1}{\cos^2 x} = \frac{3}{\sin^2 x} \]\[ \sin^2 x = 3 \cos^2 x \]. This simplifies to \( \tan^2 x = 3 \), so \( \tan x = \sqrt{3} \).
4Step 4: Solve for x
Solving \( \tan x = \sqrt{3} \) in the interval \( 0 < x < \frac{\pi}{2} \), we get \( x = \frac{\pi}{3} \). This is where the function may have an extreme value.
5Step 5: Evaluate the Function at Critical Point
Substitute \( x = \frac{\pi}{3} \) into the original function to find the y-value:\[ y = \tan\left(\frac{\pi}{3}\right) + 3 \cot\left(\frac{\pi}{3}\right) = \sqrt{3} + 3 \cdot \frac{1}{\sqrt{3}} \]\[ = \sqrt{3} + 3 \cdot \frac{1}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3}. \]
6Step 6: Check for Endpoints and Other Values
Since the function is not defined at endpoints, we check only the interior critical point \( x = \frac{\pi}{3} \). No other critical points exist in \( 0 < x < \frac{\pi}{2} \).
7Step 7: Graph the Function
Graph \( y = \tan x + 3 \cot x \) over the interval \( 0 < x < \frac{\pi}{2} \). You'll observe that the function decreases to \( x = \frac{\pi}{3} \) and increases after, confirming that \( 2\sqrt{3} \) is indeed the absolute minimum value on this interval.
Key Concepts
DerivativeTrigonometric FunctionsAbsolute Minimum
Derivative
A derivative represents the rate at which a function is changing at any given point. In simpler terms, it tells us the slope of the tangent line to the curve of the function at a specific point.
To find critical points, which are candidates for local minima or maxima, we first compute the derivative of a function and then set it equal to zero. Critical points occur where this derivative is zero or undefined.
For the function \( y = \tan x + 3 \cot x \), we computed the derivative as \( y' = \sec^2 x - 3 \csc^2 x \). By setting \( y' = 0 \), we determined the equation \( \sec^2 x = 3 \csc^2 x \), which allowed us to solve for \( x \). This analysis helps identify where the function may have extreme values, such as a minimum.
To find critical points, which are candidates for local minima or maxima, we first compute the derivative of a function and then set it equal to zero. Critical points occur where this derivative is zero or undefined.
For the function \( y = \tan x + 3 \cot x \), we computed the derivative as \( y' = \sec^2 x - 3 \csc^2 x \). By setting \( y' = 0 \), we determined the equation \( \sec^2 x = 3 \csc^2 x \), which allowed us to solve for \( x \). This analysis helps identify where the function may have extreme values, such as a minimum.
Trigonometric Functions
Trigonometric functions, like tangent (\( \tan x \)) and cotangent (\( \cot x \)), are vital in trigonometry.
They relate angles of a triangle to the lengths of its sides and have properties that make them periodic and oscillatory.
The functions \( \tan x \) and \( \cot x \) are particularly interesting because they become undefined at certain points. For example, \( \tan x \) is undefined at \( x = 0 \) and \( x = \frac{\pi}{2} \), while \( \cot x \) is undefined at multiples of \( \pi \).
In our problem, we encounter both \( \tan \) and \( \cot \) functions, which contribute to the behavior of the combined function \( y = \tan x + 3 \cot x \) within the interval \( 0 < x < \frac{\pi}{2} \). Trigonometric functions often involve derivatives like \( \sec^2 x \) for \( \tan x \) and \( -\csc^2 x \) for \( \cot x \), which are crucial to finding critical points and understanding their growth and decay on a given interval.
They relate angles of a triangle to the lengths of its sides and have properties that make them periodic and oscillatory.
The functions \( \tan x \) and \( \cot x \) are particularly interesting because they become undefined at certain points. For example, \( \tan x \) is undefined at \( x = 0 \) and \( x = \frac{\pi}{2} \), while \( \cot x \) is undefined at multiples of \( \pi \).
In our problem, we encounter both \( \tan \) and \( \cot \) functions, which contribute to the behavior of the combined function \( y = \tan x + 3 \cot x \) within the interval \( 0 < x < \frac{\pi}{2} \). Trigonometric functions often involve derivatives like \( \sec^2 x \) for \( \tan x \) and \( -\csc^2 x \) for \( \cot x \), which are crucial to finding critical points and understanding their growth and decay on a given interval.
Absolute Minimum
The absolute minimum refers to the smallest value a function attains over a specified interval or domain.
Finding this value involves evaluating the function at its critical points, as well as considering the endpoints of the interval.
However, in our problem, \( y = \tan x + 3 \cot x \) is not defined at the endpoints \( 0 \) and \( \frac{\pi}{2} \). Therefore, the critical point \( x = \frac{\pi}{3} \) becomes the main candidate for the absolute minimum value.
By substituting back into the original function, we found \( y \) to be \( 2\sqrt{3} \) at this point. Since no other critical points exist in the interval \( 0 < x < \frac{\pi}{2} \), \( 2\sqrt{3} \) is confirmed as the absolute minimum value of \( y \) on the specified interval.
Understanding absolute minima is important in various fields, as they often represent optimal or least-cost solutions.
Finding this value involves evaluating the function at its critical points, as well as considering the endpoints of the interval.
However, in our problem, \( y = \tan x + 3 \cot x \) is not defined at the endpoints \( 0 \) and \( \frac{\pi}{2} \). Therefore, the critical point \( x = \frac{\pi}{3} \) becomes the main candidate for the absolute minimum value.
By substituting back into the original function, we found \( y \) to be \( 2\sqrt{3} \) at this point. Since no other critical points exist in the interval \( 0 < x < \frac{\pi}{2} \), \( 2\sqrt{3} \) is confirmed as the absolute minimum value of \( y \) on the specified interval.
Understanding absolute minima is important in various fields, as they often represent optimal or least-cost solutions.
Other exercises in this chapter
Problem 69
\begin{equation} \begin{array}{l}{\text { a. The function } y=\cot x-\sqrt{2} \csc x \text { has an absolute maxi- }} \\ {\text { mum value on the interval } 0
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