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Chapter 4: Problem 41

Find the function's absolute maximum and minimum values and say where they occur. $$f(x)=x^{4 / 3}, \quad-1 \leq x \leq 8$$

Short Answer

Expert verified
Absolute maximum is 16 at \(x = 8\) and absolute minimum is 0 at \(x = 0\).

Step by step solution

01

Determine the critical points

To find the critical points, first differentiate the function with respect to \(x\):\[ f'(x) = \frac{4}{3}x^{1/3} \]Set the derivative equal to zero to find critical points:\[ \frac{4}{3}x^{1/3} = 0 \]Solving this gives:\[ x = 0 \] as the critical point. Check that this point lies within the interval \(-1 \leq x \leq 8\). It does.
02

Evaluate the function at endpoints and critical points

Evaluate the function \(f(x) = x^{4/3}\) at the endpoints of the interval and the critical point.- At \(x = -1\), \(f(-1) = (-1)^{4/3} = 1\).- At \(x = 0\), \(f(0) = 0^{4/3} = 0\).- At \(x = 8\), \(f(8) = 8^{4/3} = (2^3)^{4/3} = 2^4 = 16\).
03

Compare function values

The possible values from Step 2 are:- \(f(-1) = 1\)- \(f(0) = 0\)- \(f(8) = 16\)The absolute maximum value of the function is 16 at \(x = 8\). The absolute minimum value of the function is 0 at \(x = 0\).

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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 is crucial in calculus as they help us determine where a function changes behavior. In the given exercise, finding the critical points of the function \(f(x) = x^{4/3}\) involves differentiation, which is the first step. Critical points occur where the derivative is either zero or undefined.

To find these points, differentiate the function with respect to \(x\). This gives us the derivative \(f'(x) = \frac{4}{3}x^{1/3}\).
  • Set \(f'(x)\) equal to zero: \(\frac{4}{3}x^{1/3} = 0\).
  • Solve for \(x\). For this function, \(x = 0\) is a critical point.
This critical point, \(x = 0\), lies within the interval \([-1, 8]\). In other problems, be sure to check that the critical point is within the given domain.
Maximum and Minimum Values
The search for maximum and minimum values, or extrema, helps in various real-world applications. To find these values on the given interval \( -1 \leq x \leq 8 \), focus on the endpoints and critical points. These points are where the function could have its highest or lowest values.
  • For the function \(f(x) = x^{4/3}\), evaluate it at its critical point and endpoints: \(-1, 0,\) and \(8\).
  • Calculate:
    • At \(x = -1\), \(f(-1) = 1\).
    • At \(x = 0\), \(f(0) = 0\).
    • At \(x = 8\), \(f(8) = 16\).
Comparing these values reveals the absolute maximum and minimum values of the function on the specified interval.

- Absolute maximum value: \(16\) at \(x = 8\)
- Absolute minimum value: \(0\) at \(x = 0\)
Differentiation
Differentiation is a key operation in calculus, used to find the rate of change or the slope of a function. It turns a function into its derivative, which can provide insightful information about the behavior of the original function.

In this exercise with \(f(x) = x^{4/3}\), differentiation gives us the derivative \(f'(x) = \frac{4}{3}x^{1/3}\). This tells us how \(f(x)\) changes as \(x\) changes. Differentiation, in this context, helps identify critical points by showing where the derivative equals zero or is undefined.
  • Derivatives can indicate increasing or decreasing behavior in a function.
  • They also help calculate the slope at any point of the original function.
  • By setting the derivative to zero, we locate potential extrema.
Understanding how to differentiate and use derivatives can make tackling calculus problems like this one much more intuitive.

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Most popular questions from this chapter

Solve the initial value problems. $$\frac{d^{3} \theta}{d t^{3}}=0 ; \quad \theta^{\prime \prime}(0)=-2, \quad \theta^{\prime}(0)=-\frac{1}{2}, \quad \theta(0)=\sqrt{2}$$

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Right, or wrong? Say which for each formula and give a brief reason for each answer. a. \(\int \tan \theta \sec ^{2} \theta d \theta=\frac{\sec ^{3} \theta}{3}+C\) b. \(\int \tan \theta \sec ^{2} \theta d \theta=\frac{1}{2} \tan ^{2} \theta+C\) c. \(\int \tan \theta \sec ^{2} \theta d \theta=\frac{1}{2} \sec ^{2} \theta+C\)

a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\).

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int(2 \cos 2 x-3 \sin 3 x) d x$$
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