Problem 70 Sketch the graph of the given fu... [FREE SOLUTION]

Chapter 2: Problem 70

Sketch the graph of the given function $f$ on the interval [-1.3,1.3] $$ f(x)=x^{4}+2 $$

Short Answer

Expert verified

The graph of the function $f(x) = x^4 + 2$ on the interval $[-1.3, 1.3]$ has a local minimum at the point $(0, 2)$ and is decreasing on the interval $-1.3 < x < 0$ and increasing on the interval $0 < x < 1.3$. To sketch the graph, mark the local minimum and plot the ends of the interval, then draw the curve decreasing to the left of the minimum and increasing to the right.

Step by step solution

Find the derivative of the function

First, determine the derivative of the given function with respect to x. This will help us determine the intervals on which it is increasing or decreasing. \[ f'(x) = \frac{d}{dx}(x^4 + 2) = 4x^3 \]

Identify critical points

Next, find the critical points of the function by finding the values of x for which $f'(x) = 0$ or is undefined. In this case, since the derivative is a polynomial, we only need to find the values of x for which the derivative is 0. \[ 4x^3 = 0 \Rightarrow x = 0 \] So, there is a critical point at $x = 0$.

Determine increasing and decreasing intervals

To find the increasing and decreasing intervals of the function, we will examine the sign of the derivative. Divide the number line into intervals using the critical points found in the previous step. Interval 1: $x < 0$ For this interval, plug in any number less than 0, for example, $x = -1$. \[ f'(-1) = -4 \] Since $f'(-1) < 0$, the function is decreasing on the interval $-1.3 < x < 0$. Interval 2: $x > 0$ For this interval, plug in any number greater than 0, for example, $x = 1$. \[ f'(1) = 4 \] Since $f'(1) > 0$, the function is increasing on the interval $0 < x < 1.3$.

Determine local extrema

Since the function changes from decreasing to increasing at $x = 0$, there is a local minimum at that point. To find the value of the function at the local minimum, plug in $x = 0$. \[ f(0) = 0^4 + 2 = 2 \] So, there is a local minimum at the point $(0, 2)$.

Sketch the graph

With all this information, we can now sketch the graph of the function $f(x) = x^4 + 2$ on the interval $[-1.3, 1.3]$. Mark the local minimum at $(0, 2)$, draw the curve decreasing to the left of the minimum and increasing to the right, and plot the ends of the interval to complete the sketch. Remember that this is just a sketch and may not be accurate to scale; however, it should reflect the general behavior of the function on the given interval.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Critical Points

Critical points are crucial in defining where a function changes its behavior. They are the points on the graph where the derivative is zero or undefined. For the function $ f(x) = x^4 + 2 $, its derivative is $ f'(x) = 4x^3 $. Since this derivative is a polynomial, it is defined everywhere, so there are no points where it is undefined. To find when the derivative is zero, solve the equation:

$ f'(x) = 4x^3 = 0 $
Solve for $ x $: $ x = 0 $

This means the critical point occurs at $ x = 0 $. It's important because it might indicate local extrema or a change from increasing to decreasing intervals or vice versa.

Local Extrema

Local extrema come in two flavors: local minimums and local maximums. These points represent where a function reaches a local highest or lowest point relative to its nearby values. In our case, for the function $ f(x) = x^4 + 2 $, the critical point at $ x = 0 $ determines if we have an extremum. By checking the derivative, we know:

Before $ x = 0 $, the function decreases since $ f'(x) < 0 $.
After $ x = 0 $, it increases because $ f'(x) > 0 $.

This change from decreasing to increasing indicates a local minimum at $ x = 0 $. To find the function's value at this local minimum, substitute $ x = 0 $ into the original function:

$ f(0) = 0^4 + 2 = 2 $

Therefore, the point $(0, 2)$ is where our local minimum lies.

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing involves analyzing the derivative's sign. The derivative $ f'(x) = 4x^3 $ helps determine these intervals:

For $ x < 0 $, choose a test point (e.g., $ x = -1 $). $ f'(-1) = -4 $
This indicates that the function is decreasing for $ -1.3 < x < 0 $.
For $ x > 0 $, test with a point like $ x = 1 $. $ f'(1) = 4 $
The positive value infers that the function is increasing on $ 0 < x < 1.3 $.

These intervals suggest a behavior where the function first decreases, reaching its lowest point at $ x = 0 $, and then starts increasing beyond that.

Derivative

The derivative is a powerful tool in calculus that provides the rate at which a function is changing. For the function $ f(x) = x^4 + 2 $, the derivative is calculated as $ f'(x) = 4x^3 $. This third-degree polynomial tells us vital information about the graph:

Sign of the derivative $ f'(x) $ indicates if the function is increasing or decreasing.
Positive derivative values mean the function is increasing.
Negative derivative values mean the function is decreasing.

The derivative itself was obtained using basic differentiation rules applied to $ x^4 $. The constant term, 2, disappears because constants have zero derivatives. Understanding derivatives can give us insights into the function's slope at any point, helping to precisely map the behavior of the graph.

91影视

Sketch the graph of the given function \(f\) on the interval [-1.3,1.3] $$ f(x)=x^{4}+2 $$

Short Answer

Step by step solution

Find the derivative of the function

Identify critical points

Determine increasing and decreasing intervals

Determine local extrema

Sketch the graph

Key Concepts

Critical Points

Local Extrema

Increasing and Decreasing Intervals

Derivative

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Geometry

Statistics

Theoretical and Mathematical Physics

Logic and Functions

Applied Mathematics

Study anywhere. Anytime. Across all devices.