Problem 4 - Essential Calculus

Question

Determine the open intervals on which the graph is concave upward or concave downward. \(y=\frac{-3 x^{5}+40 x^{3}+135 x}{270}\)

Step-by-Step Solution

Answer

The graph of \(y=\frac{-3 x^{5}+40 x^{3}+135 x}{270}\) is concave upward on the intervals \(-\sqrt{5}\sqrt{5}\), and concave downward on the intervals \(x<-\sqrt{5}\) and \(0

Step 1: Find the first derivative

The first derivative of \(y\) is given by \(y'=\frac{d}{dx} \left(\frac{-3 x^{5}+40 x^{3}+135 x}{270}\right)\). This simplifies to \(y'=\frac{-15x^{4}+120x^{2}+135}{270}\).

Step 2: Find the second derivative

Next, differentiate \(y'\) to find the second derivative. \(y''=\frac{d}{dx} \left(\frac{-15x^{4}+120x^{2}+135}{270}\right)\) simplifies to \(y''=\frac{-60x^{3}+240x}{270}\).

Step 3: Solve for critical points

To find the critical points of the graph, set \(y''=0\) and solve for \(x\). The solutions, \(x=0\) and \(x=\pm\sqrt{5}\), represent points where the graph might change concavity.

Step 4: Determine concavity

The graph is concave upward where \(y''>0\), and concave downward where \(y''<0\). Using the critical points, \(x<-\sqrt{5}\), \(-\sqrt{5}\sqrt{5}\) are the intervals to be tested.

Step 5: Test the intervals

Substitute a value from each interval into \(y''\) to find its sign. \(y''\) is positive for \(-\sqrt{5}\sqrt{5}\), and negative for \(x<-\sqrt{5}\) and \(0\sqrt{5}\), and concave downward on the intervals \(x<-\sqrt{5}\) and \(0

Key Concepts

First Derivative

In calculus, the first derivative of a function gives us crucial information about the function's behavior. It represents the rate of change or the slope of the tangent line at any given point on a graph.
When you calculate the first derivative of a function, you're essentially determining how the function values change as the input values change.
For the function given in the exercise, the first derivative was found to be:

• \(y' = \frac{-15x^{4} + 120x^{2} + 135}{270} \)

This derivative helps us understand where the function is increasing and decreasing. Where the first derivative is positive, the function is increasing; where it's negative, the function is decreasing. However, to determine concavity, we must dig deeper by taking the second derivative. Let's uncover what that means.

Second Derivative

The second derivative of a function provides insights into the concavity of the function's graph. Concavity essentially tells us the direction in which a function is "curving."
If a graph is "concave upward," it resembles the shape of a cup and is often called "convex." Conversely, if it is "concave downward," it resembles an upside-down cup.In this case, the second derivative is given by:

• \(y'' = \frac{-60x^{3} + 240x}{270} \)

By evaluating this derivative, we can determine which intervals of the function are concave upward and which are concave downward. When the second derivative is positive, the graph is concave upward. Conversely, when it’s negative, it indicates the graph is concave downward. Next, let’s explore how we find the specific points of interest by determining the critical points of this function.

Critical Points

Critical points in calculus are where the graph of a function might change its direction of concavity, shape, or curvature. To find these, we typically set the second derivative equal to zero and solve for the variable.
This is because at critical points, the concavity can transition from upward to downward or vice versa. These points can often indicate local maxima or minima, although, for assessing concavity, they primarily mark potential changes in curvature.For this problem, solving \( y'' = 0 \) results in critical points:

• \(x = 0\)
• \(x = \pm \sqrt{5} \)

These critical points hint at the location where the graph might switch its concave direction. Once you have these values, they can be used to examine surrounding intervals and ultimately determine concavity on the graph.

Functions

In mathematics, a function is a relation between a set of inputs and a set of possible outputs, typically represented as a graph. Functions are fundamental building blocks in calculus, providing a way to model real-world relationships mathematically.
The graph of a function demonstrates how, for each input value, there is a unique output or result.In the context of this exercise, we explored a function expressed in terms of x:

• \(y = \frac{-3x^{5} + 40x^{3} + 135x}{270}\)

By examining derivatives of this function, we can draw conclusions about its shape, direction of opening (concavity), and other properties. Understanding the function’s behavior helps solve many calculus problems by breaking them down into manageable parts.

Previous

Problem 3

Next

Problem 4

Other exercises in this chapter

Problem 3

Numerical and Graphical Analysis In Exercises 3-8, use a graphing utility to complete the table and estimate the limit as \(x\) approaches infinity. Then use a

View solution

Problem 3

Identify the open intervals on which the function is increasing or decreasing. $$ f(x)=x^{2}-6 x+8 $$

View solution

Problem 4

Find two positive numbers that satisfy the given requirements. The product is 192 and the sum is a minimum.

View solution

Problem 4

Numerical and Graphical Analysis In Exercises 3-8, use a graphing utility to complete the table and estimate the limit as \(x\) approaches infinity. Then use a

View solution

Key Concepts

First Derivative

In calculus, the first derivative of a function gives us crucial information about the function's behavior. It represents the rate of change or the slope of the tangent line at any given point on a graph.
When you calculate the first derivative of a function, you're essentially determining how the function values change as the input values change.
For the function given in the exercise, the first derivative was found to be:

• \(y' = \frac{-15x^{4} + 120x^{2} + 135}{270} \)

This derivative helps us understand where the function is increasing and decreasing. Where the first derivative is positive, the function is increasing; where it's negative, the function is decreasing. However, to determine concavity, we must dig deeper by taking the second derivative. Let's uncover what that means.

Second Derivative

The second derivative of a function provides insights into the concavity of the function's graph. Concavity essentially tells us the direction in which a function is "curving."
If a graph is "concave upward," it resembles the shape of a cup and is often called "convex." Conversely, if it is "concave downward," it resembles an upside-down cup.In this case, the second derivative is given by:

• \(y'' = \frac{-60x^{3} + 240x}{270} \)

By evaluating this derivative, we can determine which intervals of the function are concave upward and which are concave downward. When the second derivative is positive, the graph is concave upward. Conversely, when it’s negative, it indicates the graph is concave downward. Next, let’s explore how we find the specific points of interest by determining the critical points of this function.

Critical Points

Critical points in calculus are where the graph of a function might change its direction of concavity, shape, or curvature. To find these, we typically set the second derivative equal to zero and solve for the variable.
This is because at critical points, the concavity can transition from upward to downward or vice versa. These points can often indicate local maxima or minima, although, for assessing concavity, they primarily mark potential changes in curvature.For this problem, solving \( y'' = 0 \) results in critical points:

• \(x = 0\)
• \(x = \pm \sqrt{5} \)

These critical points hint at the location where the graph might switch its concave direction. Once you have these values, they can be used to examine surrounding intervals and ultimately determine concavity on the graph.

Functions

In mathematics, a function is a relation between a set of inputs and a set of possible outputs, typically represented as a graph. Functions are fundamental building blocks in calculus, providing a way to model real-world relationships mathematically.
The graph of a function demonstrates how, for each input value, there is a unique output or result.In the context of this exercise, we explored a function expressed in terms of x:

• \(y = \frac{-3x^{5} + 40x^{3} + 135x}{270}\)

By examining derivatives of this function, we can draw conclusions about its shape, direction of opening (concavity), and other properties. Understanding the function’s behavior helps solve many calculus problems by breaking them down into manageable parts.