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Chapter 3: Problem 357

=Consider the function \(y=x^{1 / x}\) for \(x > 0\) a. Determine the points on the graph where the tangent line is horizontal. b. Determine the points on the graph where \(y^{\prime} > 0\) and those where \(y^{\prime} < 0\)

Short Answer

Expert verified
a. Tangent line is horizontal at \((e, e^{1/e})\). b. \(y' > 0\) for \(x < e\), \(y' < 0\) for \(x > e\).

Step by step solution

01

Understand the Problem

The function given is \(y = x^{1/x}\). We need to find points where the tangent line is horizontal, which occurs where \(y' = 0\). Additionally, we need to understand where \(y' > 0\) and \(y' < 0\). To begin, we have to find the derivative \(y'\) of the function.
02

Express Function in a More Convenient Form

To differentiate easily, let's rewrite \(y = x^{1/x}\) using logarithms. Take the natural logarithm of both sides: \(\ln(y) = \frac{1}{x}\ln(x)\). We'll derive \(y\) implicitly.
03

Implicit Differentiation

Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}\left(\frac{\ln(x)}{x}\right) \]This results in \[ \frac{1}{y}y' = \frac{1 - \ln(x)}{x^2} \] Hence, \[ y' = y \cdot \frac{1 - \ln(x)}{x^2} \] Substitute back \(y = x^{1/x}\): \[ y' = x^{1/x} \cdot \frac{1 - \ln(x)}{x^2} \].
04

Find Where the Derivative is Zero

The derivative \(y'\) is zero when the numerator is zero. Thus, solve \[ 1 - \ln(x) = 0 \] which gives \( \ln(x) = 1 \). Hence, \( x = e \). The point on the graph is \( (e, e^{1/e}) \).
05

Determine Sign of the Derivative

The derivative \(y' = x^{1/x} \cdot \frac{1 - \ln(x)}{x^2}\) changes sign based on \(1 - \ln(x)\):- For \(1 - \ln(x) > 0\), we find \(x < e\), thus \(y' > 0\).- For \(1 - \ln(x) < 0\), we find \(x > e\), thus \(y' < 0\).

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Key Concepts

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

Derivative of a function
In calculus, the derivative of a function represents how the function's value changes as its input changes. It is a fundamental concept that allows us to understand the behavior of functions.
The derivative is often denoted as \( y' \) or \( \frac{dy}{dx} \), and for any function \( y = f(x) \), it expresses the rate of change or slope at any given point \( x \).
  • For the function \( y = x^{1/x} \), our task is to find \( y' \).
  • This derivative gives us insights into whether the function is increasing or decreasing.
To decipher this, we look at the change in the function around different values of \( x \). Calculating the derivative is a key step in tackling calculus problems that involve graph analysis and understanding the nature of curves.
Implicit Differentiation
Implicit differentiation is a technique used when a function is not straightforwardly presented as \( y = f(x) \). Instead, it's buried in a more complex equation.
With implicit differentiation, we differentiate each term separately, treating \( y \) as a function of \( x \), even if it doesn鈥檛 explicitly appear that way.
  • For \( y = x^{1/x} \), we first take the natural log on both sides to rewrite it as \( \ln(y) = \frac{1}{x}\ln(x) \).
  • This helps us simplify the differentiation process.
Differentiating implicitly involves using the chain rule. For instance, differentiating \( \ln(y) \) gives \( \frac{1}{y}y' \) because we treat \( y \) as a function of \( x \). This is a powerful technique for handling complex functional forms.
Horizontal Tangent Lines
A tangent line is horizontal when its slope, given by the derivative, is zero. For a function \( y = f(x) \), this means finding \( x \) where \( y' = 0 \).
In our problem:
  • The derivative is \( y' = x^{1/x} \cdot \frac{1 - \ln(x)}{x^2} \).
  • The numerator, \( 1 - \ln(x) \), dictates when the derivative is zero.
By setting \( 1 - \ln(x) = 0 \), we solve to find \( x = e \).
At \( x = e \), the tangent line is horizontal because the slope at this point equals zero. The corresponding \( y \) value is \( e^{1/e} \), so the point on the graph with a horizontal tangent is \( (e, e^{1/e}) \).
This knowledge helps in determining peaks or troughs in the function's graph where the rate of change comes to a pause.
Graph Analysis
Graph analysis involves understanding various aspects of a function's graph, such as slope, behavior, and critical points.\( y' \), the derivative, is crucial for this.
  • The sign of \( y' \) indicates whether the function is increasing or decreasing.
  • For \( y = x^{1/x} \), \( y' > 0 \) when \( x < e \), implying the function is increasing.
  • Conversely, \( y' < 0 \) for \( x > e \), suggesting the function is decreasing.
Through graph analysis, we can identify critical transition points. When the graph moves from increasing to decreasing, these are often points of local maxima or minima.
In our case, critical analysis of the graph of \( y = x^{1/x} \) around \( x = e \) reveals that the behavior shifts, providing vital insights into the function's structure and helping solve related calculus problems effectively.

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Most popular questions from this chapter

For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=\frac{10^{x}}{\ln 10}$$

IT] For the following position functions \(y=s(t), \quad\) an object is moving along a straight line, where \(t\) is in seconds and \(s\) is in meters. Find a. the simplified expression for the average velocity from \(t=2\) to \(t=2+h\) b. the average velocity between \(t=2\) and \(t=2+h, \quad\) where \((\mathrm{i}) h=0.1, \quad\) (ii) \(h=0.01\) (iii) \(h=0.001,\) and (iv) \(h=0.0001 ;\) and c. use the answer from a. to estimate the instantaneous velocity at \(t=2\) second. $$ s(t)=2 t^{3}+3 $$

For the following exercises, use the given values to find \(\left(f^{-1}\right)^{\prime}(a).\) $$f\left(\frac{1}{3}\right)=-8, f^{\prime}\left(\frac{1}{3}\right)=2, a=-8$$

\([\mathrm{T}]\) For the function \(f(x)=\frac{x^{2}}{x^{2}+1}, \quad\) do the following. a. Use a graphing calculator to graph \(f\) in an appropriate viewing window. b. Use the nDeriv function on a graphing calculator to find \(f^{\prime}(-4), f^{\prime}(-2), f^{\prime}(2),\) and \(f^{\prime}(4)\) .

For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=x^{2} e^{x}$$
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