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Chapter 3: Problem 18

Find (without using a calculator) the absolute extreme values of each function on the given interval. $$ f(x)=\sqrt[3]{x^{2}} \text { on }[-1,8] $$

Short Answer

Expert verified
The absolute minimum is 0 at \( x = 0 \), and the absolute maximum is 4 at \( x = 8 \).

Step by step solution

01

Find the derivative of the function

To find the critical points of the function, we start by finding the derivative of the function \( f(x) = \sqrt[3]{x^2} = x^{2/3} \). Using the power rule for differentiation, the derivative is \( f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}} \).
02

Find critical points

The critical points occur where the derivative is zero or undefined. The derivative \( \frac{2}{3x^{1/3}} \) is never zero, but it is undefined when the denominator is zero. This happens at \( x = 0 \). Thus, \( x = 0 \) is a critical point.
03

Evaluate the function at the endpoints

To find the absolute extremum, evaluate the function at both endpoints of the interval \([-1, 8]\) and at any critical points. Start with the endpoints:- For \( x = -1 \), \( f(-1) = \sqrt[3]{(-1)^2} = 1 \).- For \( x = 8 \), \( f(8) = \sqrt[3]{8^2} = \sqrt[3]{64} = 4 \).
04

Evaluate the function at the critical point

Evaluate the function at the critical point found in Step 2:- For \( x = 0 \), \( f(0) = \sqrt[3]{0^2} = 0 \).
05

Determine the absolute extrema

Now compare the values of \( f(x) \) evaluated at the endpoints and the critical point. We have:- \( f(-1) = 1 \)- \( f(0) = 0 \)- \( f(8) = 4 \)The smallest value is \( 0 \) at \( x = 0 \), and the largest value is \( 4 \) at \( x = 8 \). Thus, the absolute minimum is 0 at \( x = 0 \), and the absolute maximum is 4 at \( x = 8 \).

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Key Concepts

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

Understanding Critical Points
Critical points in a function are where the derivative is zero or undefined. They are crucial in finding the extrema (minimum or maximum values) of the function. For our function, \( f(x) = \sqrt[3]{x^2} \), the derivative simplifies to \( f'(x) = \frac{2}{3x^{1/3}} \). This derivative offers insights into the behavior of the function.
  • The derivative is never zero because \( \frac{2}{3x^{1/3}} \) has no numerator that solves to zero.
  • It's undefined where the denominator is zero; precisely, at \( x = 0 \). Here, the cube root of zero in the denominator makes the expression undefined.
Thus, \( x = 0 \) is the critical point. Critical points help us decide where a function may reach its highest or lowest values.
Grasping the Derivative
The derivative of a function represents the rate at which the function's value changes. Using the power rule, we identified the derivative of \( f(x) = \sqrt[3]{x^2} \) to be \( f'(x) = \frac{2}{3x^{1/3}} \).
  • This tells us how the function grows or decreases at any point along the curve.
  • It acts as a tool to pinpoint critical points where potential extrema might occur on an interval.
Derivatives are the backbone of function analysis, offering detailed behavior at any instant along the function's domain.
Interval Evaluation Method
Interval evaluation is about examining function values over a specified range. In the given problem, the interval is \([-1, 8]\). These endpoints bound the area in which we search for extrema.
  • First, evaluate the function at both endpoints of the interval.
  • Determine function values at any critical points within this range.
  • Compare these values to find which is the largest and smallest.
Interval evaluation ensures that the extreme values of a function within a bounded domain are thoroughly explored.
Comprehensive Function Analysis
Function analysis involves multiple steps, including finding critical points and evaluating the function.
  • Derivatives highlight critical points, showing where the function might have peaks and troughs.
  • Endpoint evaluations confirm these observations within a given interval.
  • Finally, comparing results identifies absolute extrema; the maximum and minimum values the function can take.
Through well-rounded analysis, one can determine and verify the function鈥檚 highest and lowest outcomes within certain constraints.

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Most popular questions from this chapter

\(37-40 .\) Find the equation of the tangent line to the curve at the given point using implicit differentiation. Elliptic curve \(y^{2}=x^{3}-4 x+1\) at (-2,1) (See diagram in next column.)

The number \(x\) of printer cartridges that a store will sell per week and their price \(p\) (in dollars) are related by the equation \(x^{2}=4500-5 p^{2}\). If the price is falling at the rate of \(\$ 1\) per week, find how the sales will change if the current price is \(\$ 20 .\)

A company's demand equation is \(x=\sqrt{2000-p^{2}},\) where \(p\) is the price in dollars. Find \(d p / d x\) when \(p=40\) and interpret your answer.

Suppose that you have a formula that relates the value of an investment (denoted by \(x\) ) to the day of the year (denoted by \(y\) ). State, in everyday language, what \(\frac{d y}{d x}\) and \(\frac{d x}{d y}\) would mean.

Explain why, for a (simplified) rational function, the function and its derivative will be undefined at exactly the same \(x\) -values.
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