/*! 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 28 \(25-36\) (a) Find the interva... [FREE SOLUTION] | 91Ó°ÊÓ

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\(25-36\) (a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the inforvation from parts (a)- (c) to sketch the graph. Check your work with a graphing device if you have one. $$g(x)=200+8 x^{3}+x^{4}$$

Short Answer

Expert verified
Function is increasing everywhere, inflection points at \(x = -4\) and \(x = 0\).

Step by step solution

01

Find the first derivative

To find the intervals of increase or decrease, we need the first derivative of the function. Given:\[ g(x) = 200 + 8x^3 + x^4 \]Differentiating with respect to \(x\), we get:\[ g'(x) = 24x^2 + 4x^3 \]
02

Determine critical points

Set the first derivative equal to zero to find the critical points:\[ 24x^2 + 4x^3 = 0 \]Factor out the greatest common factor:\[ 4x^2(6 + x) = 0 \]Solve for \(x\): \(x = 0\) or \(x = -6\). These are the critical points.
03

Test intervals for increase/decrease

Choose test points in the intervals \((-\infty, -6), (-6, 0), \text{and} (0, \infty)\) and substitute them into the first derivative:- For \(x < -6\), choose \(x = -7\): \(g'(-7) = 4(-7)^2(-1) > 0\)- For \(-6 < x < 0\), choose \(x = -1\): \(g'(-1) = 4(-1)^2(5) > 0\)- For \(x > 0\), choose \(x = 1\): \(g'(1) = 4(1)^2(7) > 0\)The function is increasing in all intervals.
04

Find second derivative for concavity

Find the second derivative by differentiating the first derivative:\[ g'(x) = 24x^2 + 4x^3 \]\[ g''(x) = 48x + 12x^2 \]
05

Determine concavity and inflection points

Set the second derivative equal to zero to find potential inflection points:\[ 48x + 12x^2 = 0 \]Factor out the greatest common factor:\[ 12x(4 + x) = 0 \]Solve for \(x\): \(x = 0\) or \(x = -4\).Test the intervals \((-\infty, -4), (-4, 0), \text{and} (0, \infty)\) for concavity:- For \(x < -4\), test \(x = -5\): \(g''(-5) = 12(-5)(-1) > 0\), concave up.- For \(-4 < x < 0\), test \(x = -2\): \(g''(-2) = 12(-2)(2) < 0\), concave down.- For \(x > 0\), test \(x = 1\): \(g''(1) > 0\), concave up.Inflection points at \(x = -4\) and \(x = 0\).
06

Summarize findings for graph sketch

The function is increasing everywhere. It is concave up on \((-\infty, -4)\) and \((0, \infty)\), and concave down on \((-4, 0)\). Inflection points at \(x = -4\) and \(x = 0\). Use this information to sketch the graph.

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Key Concepts

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

First Derivative Test
The First Derivative Test is instrumental in understanding the behavior of a function in terms of its increasing and decreasing intervals. When we take the first derivative, denoted as \( g'(x) \), it provides the slope of the tangent to the function at any point \( x \). The sign of the first derivative tells us about the function's increase or decrease:
  • If \( g'(x) > 0 \), the function is increasing in that interval.
  • If \( g'(x) < 0 \), the function is decreasing in that interval.
Critical points occur where the first derivative equals zero, \( g'(x) = 0 \), or where it is undefined. These points are potential locations for local maxima or minima.To determine where a function increases or decreases, you test values in intervals around the critical points, ensuring a clear understanding of the function's behavior over different sections of the graph.
Second Derivative Test
The Second Derivative Test is a powerful tool for determining the concavity of a function and for identifying potential local maxima and minima more precisely. After finding the first derivative, the second derivative, denoted as \( g''(x) \), reveals the function's concavity:
  • If \( g''(x) > 0 \), the function is concave up (shaped like a cup) at that interval.
  • If \( g''(x) < 0 \), the function is concave down (shaped like a cap) at that interval.
Additionally, if a critical point is plugged into the second derivative:
  • If \( g''(x) > 0 \) at a critical point, it indicates a local minimum.
  • If \( g''(x) < 0 \) at a critical point, it indicates a local maximum.
The second derivative is also vital for finding inflection points, where the function changes its concavity. This occurs when \( g''(x) = 0 \) and indicates where the graph's curve changes direction.
Concavity and Inflection Points
Concavity is an important feature of a function that describes how it curves. Determining concavity requires analyzing the second derivative, \( g''(x) \):- **Concave Up Intervals:** Where \( g''(x) > 0 \). This means the function is curving upwards, resembling the shape of a U.- **Concave Down Intervals:** Where \( g''(x) < 0 \). In these intervals, the function curves downwards, like an upside-down U.Inflection points are crucial because they signal where the concavity of the function changes. These occur at points where \( g''(x) = 0 \). The transition from concave up to concave down (or vice versa) often happens at inflection points. However, verifying the change in sign around these points is essential to confirm whether a true inflection point exists.Understanding concavity and inflection points assists in sketching accurate graphs and in predicting the function's overall shape.
Critical Points
Critical points are key features in calculus that occur where the first derivative, \( g'(x) \), equals zero or is undefined. These points are potential sites of local maxima, minima, or points of inflection.Finding critical points involves setting the first derivative to zero and solving for \( x \). For instance, in the function \( g(x) = 200 + 8x^3 + x^4 \), the critical points were derived from \( 24x^2 + 4x^3 = 0 \). Solving gives \( x = 0 \) and \( x = -6 \) as critical points.Analyzing these points with the first and second derivative tests confirms their nature:
  • A first derivative around these points determines whether the function increases or decreases before and after them.
  • A second derivative confirms if these points are convex (local minima) or concave (local maxima).
Overall, critical points are essential in identifying the structure and prominent peaks or valleys on the graph of the function, offering insight into the function's behavior at specific points.

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