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Chapter 5: Problem 16

Given \(f(x)=x^{r}-r x+k\), where \(r>0\) and \(r \neq 1\), prove that (a) if \(01, f\) has a relative minimum value at \(1 .\)

Short Answer

Expert verified
If \(0 < r < 1\), \(f(x)\) has a relative maximum at \(x = 1\). If \(r > 1\), \(f(x)\) has a relative minimum at \(x = 1\).

Step by step solution

01

Find the first derivative of the function

To determine the critical points, the first derivative of the function is needed. The function given is \[ f(x) = x^r - r x + k \] Differentiate this with respect to x: \[ f'(x) = r x^{r-1} - r \]
02

Set the first derivative to zero

Find the critical points by setting the first derivative to zero and solving for x: \[ r x^{r-1} - r = 0 \] \[ r (x^{r-1} - 1) = 0 \]\[ x^{r-1} = 1 \]\[ x = 1 \]
03

Find the second derivative of the function

Now, differentiate \(f'(x)\) to find the second derivative:\[ f''(x) = r (r-1) x^{r-2} \]
04

Analyze the second derivative at the critical point

Evaluate the second derivative at the critical point \(x = 1\): \[ f''(1) = r (r-1) \]Analyze the value based on different ranges of \(r\): (a) If \(0 < r < 1\): \[ f''(1) = r (r-1) < 0 \] (b) If \(r > 1\): \[ f''(1) = r (r-1) > 0 \]
05

Draw conclusions based on the second derivative test

Using the second derivative test: (a) If \(0 < r < 1\), \(f''(1) < 0\), so \(f(x)\) has a relative maximum at \(x = 1\). (b) If \(r > 1\), \(f''(1) > 0\), so \(f(x)\) has a relative minimum at \(x = 1\).

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Key Concepts

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

derivatives
Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. In simpler terms, the derivative of a function gives us the slope of the function at any point.
To find the derivative of a function, we differentiate it with respect to the variable. For example, given the function \(f(x) = x^r - r x + k\), its first derivative is found as follows:\[ f'(x) = r x^{r-1} - r \] This tells us how the function \(f(x)\) changes with changes in \(x\).
Derivatives are crucial for finding critical points and analyzing the behavior of functions.
critical points
Critical points are values of \(x\) where the first derivative of a function is zero or undefined. These points are important because they can indicate local maxima, minima, or inflection points.
To find the critical points of the function \(f(x) = x^r - r x + k\), we set its first derivative to zero and solve for \(x\):
\[ f'(x) = r x^{r-1} - r = 0 \] \[ r (x^{r-1} - 1) = 0 \] \[ x^{r-1} = 1 \] \[ x = 1 \] Here, \(x = 1\) is the critical point of the function. This point is where the function has either a relative maximum or minimum, depending on the second derivative test.
second derivative test
The second derivative test helps us determine the nature of critical points. It involves finding the second derivative of a function and evaluating it at the critical points.
For the function \(f(x) = x^r - r x + k\), the second derivative is found as follows:
\[ f''(x) = r (r-1) x^{r-2} \] Evaluating the second derivative at the critical point \(x = 1\) gives us: \[ f''(1) = r (r-1) \] Now, we can analyze this based on the value of \(r\):
    \ \item If \(0 < r < 1\),\[ f''(1) = r (r-1) < 0 \] indicating a relative maximum at \(x = 1\). \ \item If \(r > 1\),\[ f''(1) = r (r-1) > 0 \] indicating a relative minimum at \(x = 1\).
In summary, the second derivative test helps confirm whether a critical point is a maximum, minimum, or neither based on the sign of the second derivative at that point.

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Most popular questions from this chapter

\(f(x)=\frac{x}{\left(x^{2}+4\right)^{3 / 2}} ;[0,+\infty)\)

The demand equation for a certain commodity produced by a monopolist is \(p=a-b x\), and the total cost, \(C(x)\) dollars, of producing \(x\) units is determined by \(C(x)=c+d x\), where \(a, b, c\), and \(d\) are positive constants. If the government levies a tax on the monopolist of \(t\) dollars per unit produced, show that in order for the monopolist to maximize his profits he should pass on to the consumer only one-half of the tax; that is, he should increase his unit price by \(\frac{1}{2} t\) dollars.

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F(x)=\frac{x+2}{x-2} ;[-4,4]
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