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Chapter 3: Problem 43

Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable. $$ g(t)=t-2 \ln t $$

Short Answer

Expert verified
The given function \(g(t) = t - 2\ln{t}\) has a local minimum at \((2, 2-2\ln{2})\), found using the Second Derivative Test.

Step by step solution

01

1. Calculate the first derivative and find critical points

First, let's find the derivative of \(g(t)\): \[g(t) = t - 2\ln{t}\] \[g'(t) = \frac{d}{dt}(t - 2\ln{t})\] The derivatives of \(t\) and \(-2\ln{t}\) with respect to \(t\) are: \[\frac{d}{dt}(t) = 1\] \[\frac{d}{dt}(-2\ln{t}) = -2 \cdot \frac{1}{t}\] Now, we can find the first derivative of \(g(t)\): \[g'(t) = 1 - \frac{2}{t}\] To find the critical points, we should set the first derivative to zero or check for points where the derivative is undefined: \[g'(t) = 1 - \frac{2}{t} = 0\]
02

2. Solve for critical points and calculate the second derivative

Let's solve the equation obtained in the previous step for \(t\): \[1 - \frac{2}{t} = 0\] \[\frac{2}{t} = 1\] \[t = 2\] So, we have a critical point at \(t = 2\). Now, we will calculate the second derivative to perform the Second Derivative Test: \[g''(t) = \frac{d^2}{dt^2}(1 - \frac{2}{t})\] Since the derivative of \(1\) with respect to \(t\) is \(0\), we just need to find the derivative of \(-\frac{2}{t}\) with respect to \(t\): \[\frac{d}{dt}(-\frac{2}{t}) = \frac{2}{t^2}\] Thus, the second derivative is: \[g''(t) = \frac{2}{t^2}\]
03

3. Apply the Second Derivative Test

Now let's use the Second Derivative Test on our critical point at \(t = 2\): \[g''(2) = \frac{2}{(2)^2} = \frac{1}{2}\] Since at the critical point, \(g''(2) > 0\), the Second Derivative Test tells us that there is a local minimum at \(t = 2\). To find the function value at this point, we can evaluate \(g(2)\): \[g(2) = 2 - 2\ln{2}\] Therefore, the given function \(g(t) = t - 2\ln{t}\) has a local minimum at \((2, 2-2\ln{2})\).

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Key Concepts

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

Critical Points
In the context of calculus, critical points are where a function's derivative either becomes zero or is undefined. Identifying these points is crucial as they can potentially indicate where the function reaches its highest or lowest values, known as relative maxima and minima, or where the slope of the tangent line is zero.

When analyzing the function g(t) = t - 2 ln t, we first find its derivative, which leads us to an equation that, when solved, reveals the critical points. In our case, setting g'(t) = 1 - 2/t to zero gives us the critical point at t = 2. Critical points are just the starting point in determining the nature of these extrema, which is why additional tests, such as the Second Derivative Test, are often employed.
First Derivative
The first derivative of a function, often denoted as f'(x) or g'(t) in the case of our function, represents the rate of change of the function with respect to its variable. In simpler terms, it's the slope of the tangent line to the function's graph at any given point.

To find the first derivative of our function g(t), we apply the rules of differentiation to each term separately. For the natural logarithm, the derivative of -2 ln t with respect to t is -2/t. Hence, after differentiation, the overall first derivative g'(t) = 1 - 2/t indicates how the function's rate of change behaves for different values of t.
Second Derivative
Moving beyond the first derivative, the second derivative gives us information about the curvature of a function's graph. Denoted as f''(x) or g''(t), it signifies whether the function is concave up (shaped like a cup) or concave down (shaped like a cap). It also helps to identify points of inflection where the curvature changes.

In this exercise, the second derivative g''(t) = 2/t^2 is derived from the first derivative. Since this second derivative is always positive for t > 0, it suggests the graph of g(t) is concave up everywhere. This curvature insight is especially useful in conjunction with the Second Derivative Test to confirm the nature of critical points.
Local Minimum
A local minimum is a point on a function's graph where the function value is lower than at all other nearby points. It is a type of relative extremum, important in optimization problems or in finding the lowest value of a function within a certain interval.

The Second Derivative Test helps us conclude whether a critical point is a local minimum. This test involves evaluating the second derivative at the critical point. If the second derivative is positive at that point, the function indeed has a local minimum there, as is the case with our function g(t) at t = 2. To find the actual minimum value, we simply substitute the critical point back into the original function, yielding g(2) = 2 - 2ln2.

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Most popular questions from this chapter

Use Newton's method to find the point of intersection of the graphs to four decimal places of accuracy by solving the equation \(f(x)-g(x)=0 .\) Use the initial estimate \(x_{0}\) for the \(x\) -coordinate. f(x)=x^{2}, g(x)=\sin x, \quad x_{0}=1

Use Newton's method to approximate the indicated zero of the function. Continue with the iteration until two successive approximations differ by less than \(0.0001\). The zero of \(f(x)=x-\sin x-0.5\) between \(x=0\) and \(x=2 .\) Take \(x_{0}=1\).

Tracking a Submarine A submarine traveling along a path described by the equation \(y=x^{2}+1\) (both \(x\) and \(y\) are measured in miles) is being tracked by a sonobuoy (sound detector) located at the point \((3,0)\) in the figure. a. Show that the submarine is closest to the sonobuoy if \(x\) satisfies the equation \(2 x^{3}+3 x-3=0\). Hint: Minimize the square of the distance between the points \((3,0)\) and \((x, y)\). b. Use Newton's method to solve the equation \(2 x^{3}+3 x-3=0\) c. What is the distance between the submarine and the sonobuoy at the closest point of approach? I d. Plot the graph of the function \(g(x)\) giving the distance between the submarine and the sonobuoy using the viewing window \([0,1] \times[2,4]\). Then use it to verify the result that you obtained in parts (b) and (c).

Snowfall Accumulation The snowfall accumulation at Logan Airport \(t\) hr after a 33 -hr snowstorm in Boston in 1995 is given in the following table. \begin{tabular}{|l|c|c|c|c|c|c|} \hline Hour & 0 & 3 & 6 & 9 & 12 & 15 \\ \hline Inches & \(0.1\) & \(0.4\) & \(3.6\) & \(6.5\) & \(9.1\) & \(14.4\) \\ \hline \end{tabular} \begin{tabular}{|l|c|c|c|c|c|c|} \hline Hour & 18 & 21 & 24 & 27 & 30 & 33 \\ \hline Inches & \(19.5\) & 22 & \(23.6\) & \(24.8\) & \(26.6\) & 27 \\ \hline \end{tabular} By using the logistic curve-fitting capability of a graphing calculator, it can be verified that a regression model for this data is given by $$ f(t)=\frac{26.71}{1+31.74 e^{-0.24 t}} $$ where \(t\) is measured in hours, \(t=0\) corresponds to noon of February 6 , and \(f(t)\) is measured in inches. a. Plot the scatter diagram and the graph of the function \(f\) using the viewing window \([0,36] \times[0,30]\). b. How fast was the snowfall accumulating at midnight on February \(6 ?\) At noon on February 7? c. At what time during the storm was the snowfall accumulating at the greatest rate? What was the rate of accumulation?

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=x \ln x-x $$
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