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Chapter 4: Problem 39

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=\cos ^{2} x \text { on }[0, \pi]$$

Short Answer

Expert verified
b. What are the absolute extreme values of the function on this interval? c. Which graphing utility can be used to confirm these conclusions?

Step by step solution

01

Finding the first derivative

To find the critical points of the function \(f(x)\), we first need to find its first derivative. Given that \(f(x) = \cos^2(x)\), we can rewrite \(f(x)\) as \(f(x) = (\cos(x))^2\). Now, you can apply the chain rule, and the derivative of the cosine function: $$f'(x) = 2 \cdot \cos(x) \cdot (-\sin(x)) = -2\sin(x)\cos(x)$$
02

Finding the critical points

Now that we have the first derivative \(f'(x) = -2\sin(x)\cos(x)\), we can find the critical points by setting it equal to zero: $$-2\sin(x)\cos(x) = 0$$ The critical points are the solutions to this equation. The equation is satisfied when either \(\sin(x)=0\) or \(\cos(x)=0\). Solving for \(x\), we find that: $$x = 0, \frac{\pi}{2}, \pi$$ These are the critical points of \(f(x)\) on the interval \([0, \pi]\).
03

Determining the absolute extreme values

To determine the absolute extreme values, we need to evaluate \(f(x)\) at the critical points and endpoints of the interval. Thus, we have the following values to evaluate: $$f(0) = \cos^2(0) = 1$$ $$f\left(\frac{\pi}{2}\right) = \cos^2\left(\frac{\pi}{2}\right) = 0$$ $$f(\pi) = \cos^2(\pi) = 1$$ Comparing these values, we can see that the absolute minimum value of \(f(x)\) is \(0\) at \(x = \frac{\pi}{2}\) and the absolute maximum value is \(1\) at \(x = 0\) and \(x = \pi\).
04

Using a graphing utility to confirm conclusions

To confirm our results, you can use a graphing utility like Desmos, GeoGebra, or a graphing calculator to plot the function \(f(x) = \cos^2(x)\) on the interval \([0, \pi]\). You should observe that the function reaches its lowest point at \(x = \frac{\pi}{2}\) and its highest points at \(x = 0\) and \(x = \pi\), which matches our analytical conclusions.

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Key Concepts

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

Critical Points
Understanding critical points in calculus is essential for analyzing the behavior of functions. Critical points occur where the first derivative of a function is either zero or undefined. At these points, the function ceases to increase or decrease, indicating a potential local maximum, minimum, or point of inflection.

When finding critical points, as in our exercise with the function f(x) = cos2 x, we first find the derivative of the function. If f'(x) = 0, then x is a critical number. In our case, the critical points for f(x) on the interval [0, 蟺] were found to be 0, 蟺/2, 蟺 by solving the equation -2sin(x)cos(x) = 0. It's important to remember that not all critical points will lead to local extrema; they are simply the possible candidates for further investigation using other calculus tools.
Absolute Extreme Values
Absolute extreme values refer to the highest and lowest points of a function on a given interval. It's a crucial concept to understand when investigating the overall behavior of functions, particularly in real-world contexts where we might be interested in maximizing or minimizing certain outcomes.

In our exercise, the absolute minimum and maximum values are found by evaluating the function f(x) at the critical points and the endpoints of the interval. This step is vital; skipping the end points can lead to incorrect conclusions. After evaluating, we compare these values to determine that the function has an absolute minimum of 0 at x = 蟺/2 and an absolute maximum of 1 at x = 0 and x = 蟺, indicating the absolute extreme values of f(x) on the given interval.
First Derivative Test
The first derivative test is a powerful method for identifying the relative extrema of a function 鈥� that is, the local maxima and minima. By interpreting the sign changes in the first derivative f'(x) before and after the critical points, one can draw conclusions about the nature of each critical point.

In our exercise, once we have the critical points, we would look to the derivative f'(x) just before and after each critical number to see if it changes from positive to negative (indicating a local maximum) or from negative to positive (local minimum). We would need to perform this test on all the critical points, not just where the derivative is zero but also where it is undefined, if applicable to the given function. However, for f(x) = cos2x, our critical points occur at zero points of the derivative, simplifying our application of the first derivative test.
Chain Rule in Calculus
The chain rule is an essential theorem in calculus for differentiating composite functions. When a function f(x) consists of another function within it, such as g(h(x)), the chain rule allows us to find f'(x) by multiplying the derivative of the outer function by the derivative of the inner function.

Returning to the textbook problem, we applied the chain rule to differentiate f(x) = cos2x. Here, we considered cos(x) as the 'inner function' and the squaring operation as the 'outer function'. By following the chain rule, we calculated that f'(x) = 2 * cos(x) * (-sin(x)), the derivative of f(x). The ability to apply the chain rule correctly and efficiently is vital to solving problems like finding the critical points and absolute extrema of composite functions.

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Most popular questions from this chapter

Fixed points of quadratics and quartics Let \(f(x)=a x(1-x)\) where \(a\) is a real number and \(0 \leq x \leq 1\). Recall that the fixed point of a function is a value of \(x\) such that \(f(x)= x\) (Exercises \(28-31\) ). a. Without using a calculator, find the values of \(a,\) with \(0 < a \leq 4,\) such that \(f\) has a fixed point. Give the fixed point in terms of \(a\) b. Consider the polynomial \(g(x)=f(f(x)) .\) Write \(g\) in terms of \(a\) and powers of \( x .\) What is its degree? c. Graph \(g\) for \(a=2,3,\) and 4 d. Find the number and location of the fixed points of \(g\) for \(a=2,3,\) and 4 on the interval \(0 \leq x \leq 1\).

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=e^{x}\left(x^{2}-7 x-12\right)$$

Show that \(f(x)=a^{x}\) grows faster than \(g(x)=b^{x}\) as \(x \rightarrow \infty\) if \(1

Residuals and errors Approximate the root of \(f(x)=x^{10}\) at \(x=0\) using Newton's method with an initial approximation of \(x_{0}=0.5 .\) Make a table showing the first 10 approximations, the error in these approximations (which is \(\left|x_{n}-0\right|=\left|x_{n}\right|\) ), and the residual of these approximations (which is \(f\left(x_{n}\right)\) ). Comment on the relative size of the errors and the residuals and give an explanation.

A damped oscillator The displacement of a particular object as it bounces vertically up and down on a spring is given by \(y(t)=2.5 e^{-t} \cos 2 t,\) where the initial displacement is \(y(0)=2.5\) and \(y=0\) corresponds to the rest position (see figure). a. Find the time at which the object first passes the rest position, \(y=0\) b. Find the time and the displacement when the object reaches its lowest point. c. Find the time at which the object passes the rest position for the second time. d. Find the time and the displacement when the object reaches its high point for the second time.
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