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Chapter 8: Problem 43

Graph \(y=x^{1 / x}\) for \(x>0 .\) Show what happens for very small \(x\) and very large \(x\). Indicate the maximum value.

Short Answer

Expert verified
The maximum value of the function \( y = x^{1/x} \) occurs at \( x = e \) and is \( y = e^{1/e} \approx 1.445 \).

Step by step solution

01

Understanding the Function

The function given is \( y = x^{1/x} \). This means for each \( x \), we take the \( x \)-th root of \( x \). We need to understand how this behaves as \( x \) approaches small and large values.
02

Analyzing the Behavior for Small x

For very small values of \( x \) (close to 0 but positive), \( x^{1/x} \) behaves peculiarly. As \( x \to 0^+ \), \( y = x^{1/x} \) approaches 0 because the exponent \( 1/x \) becomes very large, leading to the base \( x \) becoming very small.
03

Analyzing the Behavior for Large x

For very large values of \( x \), the function \( x^{1/x} \) approaches 1. As \( x \to \infty \), the exponent \( 1/x \) approaches 0, which makes \( x^{1/x} \to e^0 = 1 \).
04

Finding the Critical Point

To find the maximum value, we take the derivative of \( y = x^{1/x} \). Rewriting the function as \( y = e^{(\ln x)/x} \), differentiate to find critical points. Set the derivative to zero to find \( x = e \), where \( y \) reaches a maximum value.
05

Calculating the Maximum Value

At \( x = e \), we calculate \( y = e^{1/e} \). This is the maximum value of the function. Numerically, it approximates to about \( 1.444667861 \).
06

Graphing the Function

Plotting the function for \( x > 0 \), you should see it starts from zero at \( x = 0 \), rises to \( 1.444667861 \) at \( x = e \), and then gradually decreases towards 1. This shows the increasing then decreasing nature of the function.

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Key Concepts

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

Function Behavior
Understanding how a function behaves for various values of its variable is essential for graphing. Let's break it down for the function \( y = x^{1/x} \).
As x gets very small (close to 0 but still positive):
  • The term \( 1/x \) becomes very large.
  • The base \( x \) becomes very small.
This causes the function \( y = x^{1/x} \) to approach zero since a very small number raised to a large power is extremely tiny.
As x becomes very large:
  • The term \( 1/x \) gets closer to zero.
  • Thus the exponent approaches zero.
In this case, \( y \) trends towards 1 because any number to the power zero is 1.
By considering both extremes, we can visualize the trend of the function as a rise to a peak before settling towards 1.
Critical Points
Critical points are where the function's derivative equals zero or fails to exist, potentially indicating maximum or minimum values.
For \( y = x^{1/x} \), we locate the critical point to find where the function peaks. Derivative analysis helps us here. By rewriting \( y \) as \( e^{(\ln x)/x} \), differentiation becomes manageable.
The derivative is set to zero:
  • Solving the derivative provides critical points鈥攈ere at \( x = e \).
  • This value signifies where the function reaches its highest point before decreasing.
Finding critical points is crucial as it tells us about the function's peak behavior and allows us to pinpoint where significant changes occur in the function's direction.
Derivative Analysis
Derivatives help us understand the rate of change of functions and find critical points.
For \( y = x^{1/x} \), we express it as \( y = e^{(\ln x)/x} \) to use differentiation efficiently.
Taking the derivative involves applying the chain rule and results in:
  • The derivative equals zero at \( x = e \), indicating a peak or valley.
  • Differentiate to understand not just when, but how, the function changes around that point.
By analyzing this, we confirm that \( x = e \) corresponds to a local maximum. The derivative tells us about increasing or decreasing tendencies of functions, helping to shape a more precise graph and capture key features of the function's behavior.

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