Question

locate the extrema for the graph of y = f(x). name and classify the extrema of the function.
13.
locate the extrema for the graph of y = f(x). name and classify the extrema of the function.
14.
locate the extrema for the graph of y = f(x). name and classify the extrema of the function.
18.
use a graphing calculator to graph each function and to determine and classify its extrema.

1. v(w)=w³ - 7w - 6

use a graphing calculator to graph each function and to determine and classify its extrema.

1. h(x)=x⁴ - 4x² - 2

determine whether the given critical point is the location of a maximum, a minimum, or a point of inflection.

1. y = x⁵, x = 0

determine whether the given critical point is the location of a maximum, a minimum, or a point of inflection.

1. y=-x² + 8x - 10, x = 4

determine whether the given critical point is the location of a maximum, a minimum, or a point of inflection.

1. y = 1/4x⁴ - 2x², x = 2
2. a function f has a relative maximum at x = 2 and a point of inflection at x=-1. find the critical points of y=-2f(x - 5)-1. describe what happens at each new critical point.
3. business the carlisle innovation company has created a new product that costs $25 per item to produce. the company has hired a marketing analyst to help it determine a selling price for the product. after collecting and analyzing data relating selling price s to yearly consumer demand d, the analyst estimates demand for the product using the equation d=-200s + 15000.

a. if yearly profit is the difference between total revenue and production costs, determine a selling price s, s≥25, that will maximize the companys yearly profit, p. (hint: p = sd - 25d)
b. what are the risks of determining a selling price using this method?

Explanation:

Step1: Recall the derivative - based method for extrema

For a function \(y = f(x)\), we find its first - derivative \(y'=f'(x)\). Critical points occur where \(f'(x)=0\) or \(f'(x)\) is undefined. Then, we use the second - derivative test \(y'' = f''(x)\) to classify the critical points. If \(f''(c)>0\) at a critical point \(x = c\), then \(f(x)\) has a local minimum at \(x = c\). If \(f''(c)<0\), then \(f(x)\) has a local maximum at \(x = c\). If \(f''(c)=0\), the second - derivative test is inconclusive.

Problem 20: \(V(w)=w^{3}-7w - 6\)

Step1: Find the first derivative

Differentiate \(V(w)\) with respect to \(w\) using the power rule \((x^{n})'=nx^{n - 1}\).
\(V'(w)=3w^{2}-7\)

Step2: Find the critical points

Set \(V'(w) = 0\):
\[ \begin{align*} 3w^{2}-7&=0\\ 3w^{2}&=7\\ w^{2}&=\frac{7}{3}\\ w&=\pm\sqrt{\frac{7}{3}}\approx\pm1.53 \end{align*} \]

Step3: Find the second derivative

Differentiate \(V'(w)\) with respect to \(w\): \(V''(w) = 6w\)

Step4: Classify the critical points

When \(w=\sqrt{\frac{7}{3}}\), \(V''(\sqrt{\frac{7}{3}})=6\sqrt{\frac{7}{3}}>0\), so \(V(w)\) has a local minimum at \(w = \sqrt{\frac{7}{3}}\).
\(V(\sqrt{\frac{7}{3}})=(\sqrt{\frac{7}{3}})^{3}-7\sqrt{\frac{7}{3}}-6=\frac{7}{3}\sqrt{\frac{7}{3}}-7\sqrt{\frac{7}{3}}-6=-\frac{14}{3}\sqrt{\frac{7}{3}}-6\approx - 14.08\)
When \(w=-\sqrt{\frac{7}{3}}\), \(V''(-\sqrt{\frac{7}{3}})=-6\sqrt{\frac{7}{3}}<0\), so \(V(w)\) has a local maximum at \(w=-\sqrt{\frac{7}{3}}\).
\(V(-\sqrt{\frac{7}{3}})=(-\sqrt{\frac{7}{3}})^{3}+7\sqrt{\frac{7}{3}}-6=-\frac{7}{3}\sqrt{\frac{7}{3}} + 7\sqrt{\frac{7}{3}}-6=\frac{14}{3}\sqrt{\frac{7}{3}}-6\approx2.08\)

Problem 22: \(h(x)=x^{4}-4x^{2}-2\)

Step1: Find the first derivative

\(h'(x)=4x^{3}-8x = 4x(x^{2}-2)=4x(x-\sqrt{2})(x + \sqrt{2})\)

Step2: Find the critical points

Set \(h'(x)=0\), then \(x = 0,x=\sqrt{2},x=-\sqrt{2}\)

Step3: Find the second derivative

\(h''(x)=12x^{2}-8\)

Step4: Classify the critical points

When \(x = 0\), \(h''(0)=-8<0\), so \(h(x)\) has a local maximum at \(x = 0\), and \(h(0)=-2\)
When \(x=\sqrt{2}\), \(h''(\sqrt{2})=12\times2 - 8=16>0\), so \(h(x)\) has a local minimum at \(x=\sqrt{2}\), and \(h(\sqrt{2})=(\sqrt{2})^{4}-4\times(\sqrt{2})^{2}-2=4 - 8 - 2=-6\)
When \(x=-\sqrt{2}\), \(h''(-\sqrt{2})=12\times2 - 8 = 16>0\), so \(h(x)\) has a local minimum at \(x=-\sqrt{2}\), and \(h(-\sqrt{2})=(\sqrt{2})^{4}-4\times(\sqrt{2})^{2}-2=-6\)

Problem 26: \(y = x^{5},x = 0\)

Step1: Find the first derivative

\(y'=5x^{4}\)
\(y'(0)=0\)

Step2: Find the second derivative

\(y'' = 20x^{3}\)
\(y''(0)=0\)
The second - derivative test is inconclusive. We can look at the behavior of the function around \(x = 0\). For \(x<0\), \(y=x^{5}<0\) and for \(x>0\), \(y=x^{5}>0\), so \(x = 0\) is a point of inflection.

Problem 27: \(y=-x^{2}+8x - 10,x = 4\)

Step1: Find the first derivative

\(y'=-2x + 8\)
Set \(y'=0\), \(-2x + 8=0\), \(x = 4\)

Step2: Find the second derivative

\(y''=-2<0\)
So \(y\) has a local maximum at \(x = 4\), and \(y(4)=-4^{2}+8\times4 - 10=-16 + 32-10 = 6\)

Problem 30: \(y=\frac{1}{4}x^{4}-2x^{2},x = 2\)

Step1: Find the first derivative

\(y'=x^{3}-4x=x(x^{2}-4)=x(x - 2)(x + 2)\)

Step2: Find the second derivative

\(y''=3x^{2}-4\)

Step3: Evaluate the second derivative at \(x = 2\)

\(y''(2)=3\times2^{2}-4=8>0\)
So \(y\) has a local minimum at \(x = 2\), and \(y(2)=\frac{1}{4}\times2^{4}-2\times2^{2}=4 - 8=-4\)

Problem 36a:

The revenue function \(R(s)=sd=s(-200s + 15000)=-200s^{2}+15000s\)
The cost function \(C(s)=25d=25(-200s + 15000)=-5000s+375000\)
The profit function \(P(s)=R(s)-C(s)=-200s^{2}+15000s-(-5000s + 375000)=-200s^{2}+20000s-375000\)

Step1: Find the first derivative of the profit function

\(P'(s)=-400s + 20000\)

Step2: Find the critical point

Set \(P'(s)=0\)
\[ \begin{align*} -400s+20000&=0\\ -400s&=-20000\\ s&=50 \end{align*} \]

Step3: Find the second derivative of the profit function

\(P''(s)=-400<0\)
So the selling price \(s = 50\) maximizes the profit.

Problem 36b:

The risks of determining a selling price using this method include:
1. The demand function \(d=-200s + 15000\) is an estimate. In reality, the relationship between price and demand may not be linear.
2. There may be unforeseen factors such as competition, changes in consumer preferences, and economic conditions that can affect the actual demand and cost.
3. Cost estimates may be inaccurate. There could be fluctuations in raw - material prices, production inefficiencies, etc.

Answer:

For problem 20: Local minimum at \(w=\sqrt{\frac{7}{3}}\approx1.53\), \(V(\sqrt{\frac{7}{3}})\approx - 14.08\); local maximum at \(w=-\sqrt{\frac{7}{3}}\approx - 1.53\), \(V(-\sqrt{\frac{7}{3}})\approx2.08\)
For problem 22: Local maximum at \(x = 0\), \(h(0)=-2\); local minima at \(x=\pm\sqrt{2}\), \(h(\pm\sqrt{2})=-6\)
For problem 26: Point of inflection at \(x = 0\)
For problem 27: Local maximum at \(x = 4\), \(y(4)=6\)
For problem 30: Local minimum at \(x = 2\), \(y(2)=-4\)
For problem 36a: Selling price \(s = 50\) maximizes the profit
For problem 36b: Risks include inaccurate demand - price relationship, unforeseen external factors, and inaccurate cost estimates.