/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Find all relative extrema. Use t... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 23

Find all relative extrema. Use the Second Derivative Test where applicable. \(g(x)=x^{2}(6-x)^{3}\)

Short Answer

Expert verified
The given function \(g(x) = x^{2}(6-x)^{3}\) has a relative minimum at \(x=0\). The result at \(x=6\) from the Second Derivative Test was inconclusive.

Step by step solution

01

Find the First Derivative of the Function

We first need to find the derivative of \(g(x)\). To do this, we will use the Product Rule which states that \((uv)' = u'v + uv'\). Here, \(u=x^{2}\) and \(v=(6-x)^{3}\), so \(g'(x) = 2x(6-x)^{3} + x^{2}\cdot 3(6-x)^{2}\cdot (-1) = 2x(6-x)^{3} - 3x^{2}(6-x)^{2}\).
02

Find the Critical Points

Critical points occur where the derivative is zero or undefined. In this case, \(g'(x)\) is only undefined at \(x = 6\), but we can also set \(g'(x) = 0\) to find any other potential critical points. Doing so gives us \(2x(6-x)^{3} - 3x^{2}(6-x)^{2} = 0\). Simplifying this yields \(x = 0, 6\).
03

Find the Second Derivative of the Function

Now that we have our critical points, we need to find the second derivative of \(g(x)\) to be able to use the Second Derivative Test. Just like we used the Product Rule to calculate the first derivative, we'll do so again to calculate the second derivative, \(g''(x)\). The second derivative in this case is a little bit complicated, but it simplifies to \(g''(x) = 12(6-x)^{2} - 28x(6-x)\).
04

Use the Second Derivative Test

Next, we use the Second Derivative Test, which says that if a function's second derivative at a point is positive, the function has a relative minimum there, if it's negative, the function has a relative maximum there. Evaluating \(g''(x)\) at \(x=0\), we get \(12(6)^{2} = 432 > 0\), so \(x=0\) is a relative minimum. Evaluating \(g''(x)\) at \(x=6\), we get \(0\), which is neither positive nor negative, so the test is inconclusive at this point.

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Most popular questions from this chapter

Conjecture Consider the function \(f(x)=(x-2)^{n}\). (a) Use a graphing utility to graph \(f\) for \(n=1,2,3,\) and \(4 .\) Use the graphs to make a conjecture about the relationship between \(n\) and any inflection points of the graph of \(f\). (b) Verify your conjecture in part (a).

The deflection \(D\) of a beam of length \(L\) is \(D=2 x^{4}-5 L x^{3}+3 L^{2} x^{2},\) where \(x\) is the distance from one end of the beam. Find the value of \(x\) that yields the maximum deflection.

The function \(f\) is differentiable on the interval [-1,1] . The table shows the values of \(f^{\prime}\) for selected values of \(x\). Sketch the graph of \(f\), approximate the critical numbers, and identify the relative extrema. $$\begin{array}{|l|c|c|c|c|} \hline x & -1 & -0.75 & -0.50 & -0.25 \\ \hline f^{\prime}(x) & -10 & -3.2 & -0.5 & 0.8 \\ \hline \end{array}$$ $$\begin{array}{|l|l|l|l|l|l|} \hline \boldsymbol{x} & 0 & 0.25 & 0.50 & 0.75 & 1 \\ \hline \boldsymbol{f}^{\prime}(\boldsymbol{x}) & 5.6 & 3.6 & -0.2 & -6.7 & -20.1 \\ \hline \end{array}$$

In Exercises \(101-104,\) use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow \infty} \frac{2}{\sqrt{x}}=0 $$

In Exercises 61 and 62, use a graphing utility to graph the function. Then graph the linear and quadratic approximations \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) in the same viewing window. Compare the values of \(f, P_{1},\) and \(P_{2}\) and their first derivatives at \(x=a .\) How do the approximations change as you move farther away from \(x=a\) ? \(\begin{array}{ll}\text { Function } & \frac{\text { Value of } a}{a} \\\ f(x)=2(\sin x+\cos x) & a=\frac{\pi}{4}\end{array}\)
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