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Chapter 4: Problem 37

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$f(x)=x^{1 / 3}(x+8)$$

Short Answer

Expert verified
The function increases on \((-8, 0)\) and \((0, \infty)\), decreases on \((-\infty, -8)\). Local extrema may exist at \(x = -8\).

Step by step solution

01

Find the Derivative

To determine the intervals on which the function is increasing or decreasing, we first find its derivative. The given function is \( f(x) = x^{1/3}(x+8) \). Using the product rule for derivatives \((uv)' = u'v + uv'\) where \(u = x^{1/3}\) and \(v = x+8\), we get:\[ f'(x) = \frac{d}{dx}(x^{1/3}) \cdot (x+8) + x^{1/3} \cdot \frac{d}{dx}(x+8) \]Calculate \( \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{-2/3} \) and \( \frac{d}{dx}(x+8) = 1 \), so:\[ f'(x) = \frac{1}{3}x^{-2/3}(x+8) + x^{1/3}(1) \]\[ f'(x) = \frac{x+8}{3x^{2/3}} + x^{1/3} \]
02

Simplify the Derivative

Combine the terms in \( f'(x) \):Factor \( x^{-1/3} \) from each term gives:\[ f'(x) = x^{-1/3} \left( \frac{x+8}{3x^{1/3}} + x^{1/3} \right) \]\[ f'(x) = x^{-1/3} \left( \frac{x+8}{3} + x^{1/3} \right) \]Simplifying further using a common denominator gives:\[ f'(x) = \frac{1}{3}x^{-1/3}(x+8 + 3x^{1/3}) \]
03

Find Critical Points

Set \( f'(x) = 0 \) to find critical points.\[ \frac{1}{3}x^{-1/3}(x+8 + 3x^{1/3}) = 0 \]The factor \( x^{-1/3} \) implies a solution cannot occur at \( x = 0 \) because the term is undefined; check \(x+8 + 3x^{1/3} = 0\).\[ x+8 + 3x^{1/3} = 0 \]Solve for \( x \):No easy algebraic solution appears here, consider potential numerical or graphical methods to solve or bypass with estimation or testing of behavior around conjectured points.
04

Determine Increasing/Decreasing Intervals

Apply the first derivative test:Evaluate \( f'(x) \) in intervals determined by critical points and behavior around undefined point \(x = 0\).Example intervals: \( (-\infty, -8), (-8, 0), (0, +\infty) \) and test sign of \( f'(x) \) in each interval.For instance, test \( f'(-10) \), \( f'(-1) \), \( f'(1) \) for sign indication.Result signs lead to: decreasing on \( (-\infty, -8) \), increasing on \( (-8, 0) \) and \((0, +\infty) \).
05

Identify Extremes

Determine local extrema using results from step 4:Local minima/maxima occur where \( f(x) \) changes direction indicated by \( f'(x) \)'s sign change.Examine \( x = -8 \), possibly find local minimum as increasing changes to decreasing; similarly, test around other intervals.For absolute extremes, compare function values across all points of interest gathered including at boundary behaviors limit (as \(x\) approaches infinity).

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Key Concepts

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

Derivative
In calculus, a derivative is a measure of how a function changes as its input changes. Understanding the derivative is essential for analyzing the behavior of functions, such as determining intervals of increase and decrease. To find the derivative of a function like \( f(x) = x^{1/3}(x+8) \), we apply calculus rules such as the product rule. The product rule states that for two functions \( u \) and \( v \), \((uv)' = u'v + uv'\). In this exercise:
  • Let \( u = x^{1/3} \) and \( v = x+8 \).
  • The derivatives are \( u' = \frac{1}{3}x^{-2/3} \) and \( v' = 1 \).
Substituting these into the product rule gives the derivative of the function:\[ f'(x) = \frac{1}{3}x^{-2/3}(x+8) + x^{1/3} \]Understanding how to compute derivatives is foundational for exploring further critical concepts.
Critical Points
Critical points are points on the graph of a function where the derivative is zero or undefined. These points are potential locations for extreme values, such as local maxima or minima. To identify critical points, we set the derivative, \( f'(x) \), equal to zero and solve for \( x \). From the expression:\[ f'(x) = x^{-1/3} \left( \frac{x+8}{3} + x^{1/3} \right) \]we seek where the product becomes zero:
  • \( x^{-1/3} = 0 \) implies no solutions since \( x = 0 \) is undefined.
  • \( \frac{x+8}{3} + x^{1/3} = 0 \) needs solving by alternate methods such as numerical estimation, as algebraic manipulation is complex.
Thus, identifying critical points requires analyzing where changes in the derivative occur, often through testing intervals or numeric approximation rather than purely symbolic solutions.
Extreme Values
Extreme values of a function refer to its highest and lowest points, named local maxima, minima, and global maxima or minima. A function achieves local extrema at critical points where the derivative changes sign. Examining the function \( f(x) = x^{1/3}(x+8) \), observe where changes occur by analyzing the behavior of \( f'(x) \):
  • If \( f'(x) \) changes from positive to negative, \( f(x) \) has a local maximum.
  • If \( f'(x) \) changes from negative to positive, \( f(x) \) has a local minimum.
To find absolute extremes, compare function values at these critical points as well as at the endpoints of the domain, particularly considering the behavior as \( x \) approaches infinity. Determining these values establishes the overall behavior of \( f(x) \) within its domain.
Intervals of Increase and Decrease
Identifying the intervals where a function increases or decreases involves examining the sign of its derivative across different segments of its domain. For the function \( f(x) \), where\[ f'(x) = \frac{1}{3}x^{-1/3}(x+8 + 3x^{1/3}) \]we determine intervals by testing values around critical points and in distinct sections, such as:
  • When \( f'(x) > 0 \), the function is increasing. In this case, it's increasing on the interval \((-8, 0)\) and \((0, +\infty)\).
  • When \( f'(x) < 0 \), the function is decreasing. It decreases on the interval \((-\infty, -8)\).
Utilizing these intervals helps in constructing a clear picture of the function's behavior, providing insight into motion and rates of change relevant to physical and theoretical contexts.

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Most popular questions from this chapter

You are driving along a highway at a steady 60 mph \((88 \mathrm{ft} / \mathrm{sec})\) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in \(242 \mathrm{ft}\) ? To find out, carry out the following steps. 1\. Solve the initial value problem Differential equation: \(\frac{d^{2} s}{d t^{2}}=-k\) \((k \text { constant })\) Initial conditions: \(\quad \frac{d s}{d t}=88\) and \(s=0\) when \(t=0\) Measuring time and distance from when the brakes are applied 2\. Find the value of \(t\) that makes \(d s / d t=0 .\) (The answer will involve \(k .)\) 3\. Find the value of \(k\) that makes \(s=242\) for the value of \(t\) you found in Step 2

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? $$y=\frac{x^{4}}{4}-\frac{x^{3}}{3}-4 x^{2}+12 x+20$$

Consider the cubic function $$f(x)=a x^{3}+b x^{2}+c x+d$$ a. Show that \(f\) can have \(0,1,\) or 2 critical points. Give examples and graphs to support your argument. b. How many local extreme values can \(f\) have?

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=\pi x^{2} e^{-3 x / 2}, \quad[0,5]$$

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2 / 3}(x+2)$$
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