Each of Exercises 43–48 gives the first derivative of a functi...

Question

Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.
y' = 𝓍⁴ ― 2𝓍²

y' = 𝓍⁴ ― 2𝓍²

Accepted Answer

Hello. In this video, we are asked to consider the function F of X, whose first derivative is given by F X equal to X the 4th power minus 6 X2. We want to determine the X coordinates at which F of X has a local maximum, local minimum, or inflection point. So the problem gives us the first derivative, which is F of X. Equal to x of the 4th power minus 6 X 2. Now, since we want to look for both a local extrema, which is local maximums and local minimums, and inflection points, we are going to need to use the second derivative as well. So by taking the derivative of FX, FX is going to equal to 4 X cubed through the power rule minus 12 X. Let's go ahead and start by finding the possible X locations of the local maximums or minimums. So, in order to find the X coordinates for the local maximums and minimums, we want to find the locations where F of X is equal to 0. In order to do this, we will take our first derivative, which is X 4th power, minus 6 X2 and set it equal to 0. Once we solve for X, this is going to be the potential locations of a local maximum or local minimum. So, in order to solve for X, we will first start by factoring X2d from the left hand side of this equation. That will leave us with X2 multiplied by X2 minus 6 equal to 0. We can set both factors of X equal to 0, leaving us with X2 equal to 0, and X2 minus 6 equal to 0. Now, on the left hand side, if we square root both sides of the equation, we will have X's equal to 0, which is the first potential location for a local maximum or minimum value. For the second equation, we have X2 minus 6, so we will go ahead and add 6 to both sides of the equation, leaving us with X2 equal to 6. And if we take the square root, we get two more potential values, which is X is equal to positive and negative square root of 6. Now what we need to do is we need to determine which of these three values is going to be a local maximum or local minimum. In order to do this, we can go ahead and use the 2nd derivative test. Now, the second derivative test states the following. If we take a value and plug it into the second derivative, and that value is less than 0, then that implies that we have a local maximum. But if we take a value, plug it into the second derivative, and that value is greater than 0, then that implies that we have a local minimum. So what we will go ahead and do is take the 3 values we just found and plug them into the second derivative. If we take negative square root of 6 and plug this into the second derivative, F of negative square root of 6 will give us the value of -12 multiplied by the square root of 6. Then, if we take the positive square root of 6 and plug it into the second derivative, that will give us the value of 12 multiplied by square root of 6. So, notice that when we plug in negative square root of 6, we get a value that is less than 0. And when we plug in square root of 6, that gives us a value that is greater than 0. What this tells us is that X is equal. To square root of 6 will be a local maximum. And What this also tells us is that X is equal to square root of 6 will be a local minimum. So these are going to be the first two values that represent a local max or local min. But what about 0? Well, if we plug in 0 into the double derivative. That just gives us the output of 0, which doesn't really tell us anything of whether this is going to be a local max or min. So, in order to double check that 0 is a local maxim min, let's go ahead and plug this in by using the first derivative test. We will plot 0 on a number line for the first derivative test. Then we want to check the behaviors of the neighborhood of 0. In order to do this, we will go ahead and take values to the left and to the right of zero, and plug them into the first derivative. Values that we can take is -1 and positive 1. If we plug in -1 into the first derivative, the first derivative's value is going to be -5. Since the output of F-1 is -5, then that tells us that the graph will be decreasing within this region. And if we plug in 1 into our derivative, we get the value of -5 as well. That tells us that the graph is going to be decreasing to the right of 0. Now, by the first derivative test, this tells us that 0 is neither a local maximum or local minimum. So that means that the only values that represent local maxims or local minimums are going to be negative square root of 6 and square root of 6. Now let's go ahead and find inflection points. So that is going to be the second part of this problem. Now, in order to find the potential values of the inflection points, we want to find the values where the double derivative is equal to zero. So, again, our second derivative was 4 X cubed minus 12 X. So we are going to take the second derivative, 4 X cubed. Minus 12 X, set it equal to 0, and solve for the values of X. What we can start by doing is factoring out a 4 X from the left hand side of this equation. That will give us 4 X multiplied by X2 minus 3. We will set both factors of X equal to 0, so we have 4 X's equal to 0, and X squad minus 3 equal to 0. Now, on the left-hand equation, if we divide both sides of this equation by 4, that gives us X is equal to 0, which is a potential inflection point. If we add 3 and take the square root on the right-hand equation, we get X's equal to positive and negative square root of 3, which are 2 more potential inflection points. Now what we want to do is you want to check the behaviors of the concavity around these inflection points. So, we will go ahead and create a number line. We will plot our three potential inflection points on this number line, negative square root of 3. 0 and square root of 3. Then what we want to do is you want to take testing points from around the neighborhoods of these inflection points and plug them into the second derivative to determine the behaviors of the concavity within those regions. So a testing point that we can take to the left of square root of 3 will be -2. A testing point between ne square root of 3 and 0 will be -1. Between 0 and square root of 3 will be 1. And to the right of square root of 3 will be 2. So we have 4 values that we need to check. If we take -2 and plug it into the second derivative, that is going to give us the value of -8. What this tells us is that on the neighborhood from negative infinity to square root of 3, the graph is going to be concave down. If we take -1 and plug it into the second derivative, we get the value of 8. This tells us that the behavior of the graph between negative square root of 3 and 0 will be concave up. If we take the value. Of 1 and plug this into the second derivative, we will get to -8, which again tells us that this region will be concave down, and if we take the value of 2 and plug it into the second derivative, we get 8, and that tells us that the behavior of the endpoint is going to be concave up. Notice that the concavity of the graph shifts when we pass through the values of negative square root of 3. 0 and square root of 3. Since the concavity shifts when passing through these three values, then that means that all three of these values are going to be inflection points, and that is going to be the final solution to this problem. The inflection points are located at the values of X equal to -2 root of 30, and square root of 3. So I hope this video helps you in understanding on how to find these different pieces of information, and I will go ahead and see you all in the next video.

Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.y' = 𝓍⁴ ― 2𝓍²

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Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.
y' = 𝓍⁴ ― 2𝓍²