/*! 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 31 31-40. Graph each function using... [FREE SOLUTION] | 91Ó°ÊÓ

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31-40. Graph each function using a graphing calculator by first making a sign diagram for just the first derivative. Make a sketch from the screen, showing the coordinates of all relative extreme points and inflection points. Graphs may vary depending on the window chosen. $$ f(x)=x^{1 / 2} $$

Short Answer

Expert verified
The function \( f(x) = x^{1/2} \) is always increasing for \( x > 0 \), starting from the origin.

Step by step solution

01

Find the First Derivative

The first step is to find the derivative of the function given, which is \( f(x) = x^{1/2} \). Using the power rule, we get \( f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).
02

Identify Critical Points

Set the first derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) equal to zero to find critical points. Since \( \frac{1}{2\sqrt{x}} \) is undefined for \( x = 0 \), there are no points where \( f'(x) = 0 \). The critical point is at the endpoint of the domain, \( x = 0 \), since the function is not defined for \( x < 0 \).
03

Create the Sign Diagram for f'(x)

Examine the sign of \( f'(x) \) over the domain \( x > 0 \). Since \( f'(x) = \frac{1}{2\sqrt{x}} \) is always positive in this range, the function is always increasing.
04

Graph the Function

Use a graphing calculator to plot \( f(x) = x^{1/2} \). Notice how the function starts at the origin and curves upward, increasing as \( x \) increases. Since we have found the function is always increasing and no relative extrema are present, the graph should reflect this behavior.

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Key Concepts

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

First Derivative
To understand graphing functions, the concept of the first derivative plays a key role. The first derivative of a function reveals how the function is changing. For the function \( f(x) = x^{1/2} \), the first derivative is calculated using the power rule. The resulting derivative is \( f'(x) = \frac{1}{2\sqrt{x}} \). This formula allows us to determine how fast or slow the function is rising or falling.

The first derivative indicates the slope of the tangent line at any given point \( x \). A positive derivative suggests the function is increasing, while a negative derivative suggests it is decreasing. In our case, the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) is positive for all \( x > 0 \), indicating that the function is increasing across its entire domain.
Sign Diagram
Creating a sign diagram is a helpful tool to analyze the behavior of the first derivative across the function's domain. A sign diagram shows where the derivative is positive, negative, or zero.

For the function \( f(x) = x^{1/2} \), the domain is \( x > 0 \), since the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) is undefined at \( x = 0 \) and the function itself is not defined for \( x < 0 \).

In our sign diagram for \( f'(x) \), the entire domain for \( x > 0 \) shows a positive sign. This uniform positivity indicates that \( f(x) \) is consistently increasing as the value of \( x \) gets larger.
Critical Points
Critical points are locations on a graph where the derivative is either zero or undefined. These are key points because they can signify potential maxima, minima, or change in the behavior of the graph.

In the function \( f(x) = x^{1/2} \), we analyze the critical points by setting the first derivative to zero or checking where it is undefined. The derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) does not equal zero for any \( x > 0 \) but is undefined at \( x = 0 \). Since the function cannot go below zero, \( x = 0 \) acts as a boundary, contributing to it being a critical point even if there's no relative maximum or minimum at this location.
Increasing Function
An increasing function is one where the output grows larger with an increase in input. Mathematically, when the first derivative \( f'(x) > 0 \) over an interval, the function is increasing across that interval.

For \( f(x) = x^{1/2} \), since \( f'(x) = \frac{1}{2\sqrt{x}} \) is positive for the entire domain \( x > 0 \), the function is classified as increasing. This means any increase in \( x \) will result in a higher \( f(x) \) value, and graphically, the curve will rise from left to right. This characteristic shapes the entire behavior and appearance of the graph, consistently directing upwards without any dips or turns.

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Most popular questions from this chapter

GENERAL: Airplane Flight Path A plane is to take off and reach a level cruising altitude of 5 miles after a horizontal distance of 100 miles, as shown in the following diagram. Find a polynomial flight path of the form \(f(x)=a x^{3}+b x^{2}+c x+d\) by following Steps i to iv to determine the constants \(a, b, c,\) and \(d\). i. Use the fact that the plane is on the ground at \(x=\) 0 [that is, \(f(0)=0]\) to determine the value of \(d\). ii. Use the fact that the path is horizontal at \(x=0\) [that is, \(\left.f^{\prime}(0)=0\right]\) to determine the value of \(c\). iii. Use the fact that at \(x=100\) the height is 5 and the path is horizontal to determine the values of \(a\) and \(b\). State the function \(f(x)\) that you have determined. iv. Use a graphing calculator to graph your function on the window [0,100] by [0,6] to verify its shape.

\(37-40 .\) Find the equation of the tangent line to the curve at the given point using implicit differentiation. Elliptic curve \(y^{2}=x^{3}-4 x+1\) at (-2,1) (See diagram in next column.)

In each equation, \(x\) and \(y\) are functions of \(t\) Differentiate with respect to \(t\) to find a relation between \(d x / d t\) and \(d y / d t\). $$ 3 x^{2}-7 x y=12 $$

The number \(x\) of automobile tires that a factory will supply and their price \(p\) (in dollars) are related by the equation \(x^{2}=8000+5 p^{2}\). Find \(d x / d p\) at \(p=80\) and interpret your answer. [Hint: You will have to find the value of \(x\) by substituting the given value of \(p\) into the original equation.]

Write in calculus notation: The value of my car is falling at the rate of \(\$ 2000\) per year. (Be sure to define your variables.)

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