if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / ...

Question

if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.

g. Use your work in parts (a) through (f) to sketch a graph of ƒ .

Accepted Answer

Hello, in this video, we are going to be drawing the graph of the function F of X if F of X is equal to 1 divided by 6X4 plus 10. We are also told that its derivative is equal to -6X cubed divided by the quantity 3x4 + 5 squared. And we are also given the second derivative, which is 90 X 2 multiplied by X 4th minus 1, all divided by 3X4 plus 5, raised to the power of 3. So, that's a lot of information to take in, but let's go ahead and just discuss some characteristics of F of X. First, if we take a look at the original function, notice that we are given a function in the form of a rational function. So since we have a rational function, we need to go ahead and check to see if there are any restrictions in the domain. In order to do this, we want to find the values of X that may cause the denominator to be zero. So we will go ahead and take the denominator of the original function, which is 6X4th power, plus 10, set it equal to 0, and solve. In order to solve this, we will subtract 10 from both sides of the equation, leaving us with 6 X 4th power equal to -10. We divide 6 to both sides of the equation, leaving us with X the 4th power equal to -10 divided by 6. But there's going to be an issue once we take the 4th root of this equation. The problem here is that when we take the 4th root, we are going to have a negative number under an even powered square root, and that means that we are going to end up with an imaginary number, and what this indicates is that there is no restrictions in the denominator of X. So, in other words, our domain of the function is going to be all real numbers from negative infinity to positive infinity. OK, now what we want to go ahead and do is we want to go ahead and discuss the graph symmetry. So, in order to discuss the symmetry of the function, we will go ahead and plug in the value of negative X into the function and see if we get an output of the original function. So, we have F of X equal to 1 divided by 6 multiplied by negative X to the 4th power plus 10. Since negative X is raised to an even powered exponent, that means that it is going to simplify to be positive X. That will leave us with 1 divided by 6 X to the fourth power plus 10. Since this is the original function, this tells us that F of negative X is equal to F of X. This implies that the function is even, therefore, we have symmetry about the y axis. Now that we've calculated the symmetry, let's go ahead and calculate the X intercepts and Y intercepts. So, starting with the X intercepts, we want to find any values of X that cause the original function to be zero, but notice that the original function has a constant 1 in the numerator. What this means is that there are no values of X that are going to cause the function to be zero, therefore, we have no X intercepts. But what about the Y intercept? In order to calculate the Y intercept, we will need to plug in the value of 0 into our function. That will give us F of 0 is equal to 1 divided by 6 multiplied by 0 to the fourth power plus 10, and this is going to give us the value of 1/10. What this means is that we have a white intercept at the point 0.10. Now that we've discussed the X intercepts and Y intercepts, let's go ahead and calculate the critical points of the function. In order to calculate the critical points, we want to find the values of the derivative that caused the values or the values of X that cause the derivative to be zero. Now, the derivative has a numerator of 6 X cubed. We will go ahead and take 6 X cubed, set it equal to 0, and solve. So we are going to solve for critical numbers. In order to solve a critical numbers, we take the numerator of the derivative 6 X cubed, set it equal to 0, and solved. But notice that this is just going to give us X is equal to 0. So what this means is that we have a critical number around at X is equal to 0, but now we need to determine whether this is a local minimum or local maximum. In order to do this, we can check the neighborhood of 0. We will go ahead and take values around the around the 0.0. Some testing values we can pick as -1 and 1. What we want to do now is plug these values into the derivative, and based off the outputs, we will determine if the region of that graph is going to be increasing or decreasing. So, if we take the value of -1 and plug it into the derivative, the value of the derivative is going to give us 3 divided by 32, and that is greater than 0. What this means is that since the the value to the left of zero is positive for the derivative, then the original graph will be increasing within this region. Then, if we take the value of 1 and plug it into the derivative, we get the output of negative 3 divided by 32, which is less than 0. Since the derivative is negative to the right of 0, that means the original graph will be decreasing to the right of 0. And based off the first derivative test, since the graph is switching from increasing to decreasing as it passes through the value of 0, this classifies 0 to be a local maximum. So we have a local maximum at the value of 0. Now what we need to check is the inflection points, or the concavity. But in order to find concavity, we do need to find the inflection points. So, in order to calculate inflection points, We need to find the values of X that cause the second derivative to be zero. So, we will go ahead and take the numerator of the 2nd derivative, which is 90 X 2 multiplied by the quantity X the 4th minus 1, set it equal to 0, and solve for those values of X. Well, that is going to give us two values or a few values of X. We will have X is equal to 0 for the inflection point, and X is equal to positive and -1. So there are 3 inflection points in total, but now we need to check the concavity around these points. So similar to the first derivative, we will create a number line, we will go ahead and plot our possible inflection points. And we will check the concavity by checking the neighborhoods around these points. So we will take testing values around -1, 0, and 1, plug them into the second derivative, and find the values of the derivative. Some testing points will be -2, 0.5, 0.5, and 2. When we plug in -2 into the second derivative, the output that we get is a positive fraction. And since we have a positive fraction, the original graph will be concave up from negative infinity to -1. When we plug in negative 0.5 into the second derivative, we will end up with a negative value, a negative fraction, and that tells us that the graph will be concave down between -1 and 0. When we plug in 0.5 into our second derivative, we get another negative fraction, and that tells us that it is concave down when we are passing through zero, and when we plug in 2 into the second derivative, we get a positive fraction telling us that the graph is concave up within this region. So, now that we've determined the regions of increase and decrease, the inflection points. Or we have two inflection points, a negative 1 and 1, and the regions of concavity, we have the X intercept and Y intercept. We just have to choose the graph that accurately represents this information, and that graph is going to be option A. So, I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.

g. Use your work in parts (a) through (f) to sketch a graph of ƒ .

Key Concepts

Derivative

Second Derivative

Graph Sketching

Watch next