/*! 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 16 Find the local maxima and minima... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the local maxima and minima of each of the functions. Determine whether each function has local maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. \(y=(x-1)^{1 / 3}, x \geq 1\)

Short Answer

Expert verified
The function has no local maxima or minima. It is increasing on \( x > 1 \).

Step by step solution

01

Understanding Instantaneous Rate of Change

To find critical points, we must first take the derivative of the function. The function given is \( y = (x-1)^{1/3} \). Here, we need to find \( \frac{dy}{dx} \).
02

Differentiating the Function

Using the power rule for differentiation, the derivative of the function is found as \( \frac{dy}{dx} = \frac{1}{3} (x-1)^{-2/3} \). This can be rewritten as \( \frac{1}{3 \sqrt[3]{(x-1)^2}} \).
03

Finding Critical Points

Critical points occur where the derivative is zero or undefined. The derivative \( \frac{1}{3 \sqrt[3]{(x-1)^2}} \) is never zero for \( x \geq 1 \), but it is undefined at \( x = 1 \) because the denominator becomes zero.
04

Analyzing the Critical Point

The point \( x = 1 \) is the only critical point, and we need to determine if this point is a local maximum, minimum, or neither. Since the function is only defined for \( x \geq 1 \), we examine behavior as \( x \to 1^{+} \).
05

Determining Increasing/Decreasing Intervals

For \( x > 1 \), the derivative is positive, \( \frac{1}{3\sqrt[3]{(x-1)^2}} > 0 \), which means the function is increasing on \( x > 1 \). Thus, \( x = 1 \) is a point of inflection, rather than a local maximum or minimum.

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Key Concepts

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

Derivative
The derivative represents the rate at which a function is changing at any given point. In calculus, derivatives are essential to understand how a function behaves. If you imagine a graph, the derivative at a point gives you the slope of the tangent line at that point.
For the function \( y = (x-1)^{1/3} \), the derivative tells us how quickly \( y \) changes as \( x \) changes. To find the derivative \( \frac{dy}{dx} \), we use the power rule. This rule states that if a function is in the form \( y = x^n \), then its derivative is \( nx^{n-1} \).
  • Applying the power rule here, the derivative is \( \frac{1}{3} (x-1)^{-2/3} \).
  • This helps us find out where the function may have special points such as critical points – where the function behavior changes notably.
Critical Points
Critical points of a function occur where the function's derivative is either zero or undefined. At these points, a function might change direction, leading to potential peaks or valleys. For the function \( y = (x-1)^{1/3} \):
  • The derivative \( \frac{1}{3\sqrt[3]{(x-1)^2}} \) is undefined at \( x = 1 \). This is because the denominator becomes zero there.
  • As the derivative is never zero for \( x \geq 1 \), no other critical points exist aside from where the derivative is undefined.
Identifying these points is crucial as they might indicate potential local maxima or minima.
Increasing and Decreasing Intervals
The derivative also helps identify intervals where the function is increasing or decreasing.
For an interval where the derivative is positive, the function is increasing; where it is negative, the function is decreasing.
For example:
  • With \( y = (x-1)^{1/3} \), our derivative is positive for \( x > 1 \), indicating the function is increasing anytime \( x \) is greater than 1.
  • This means as \( x \) continues to grow beyond 1, \( y \) steadily increases.
This analysis shows how the function behaves over its domain and helps locate potential extremum points.
Local Maxima and Minima
Local maxima and minima are points where a function reaches a peak or trough, respectively, when examined over a small interval. However, with \( y = (x-1)^{1/3} \) for \( x \geq 1 \):
  • The critical point at \( x = 1 \) is where the derivative is undefined.
  • By observing the sign of the derivative around this point, we see only one side is defined (\( x > 1 \), it’s increasing).
  • As a result, \( x = 1 \) isn't a local maximum or minimum but rather a point of inflection where the curve sharply changes direction.
Understanding these points reveals how the function graph progresses and assists in predicting behavior over its domain.

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