In Exercises 49–52, graph each function. Then use the function...

Question

In Exercises 49–52, graph each function. Then use the function’s first derivative to explain what you see.

y = 𝓍²/³ + (𝓍―1)²/³

Accepted Answer

Using the 1st and 2nd derivatives, determine the intervals of increase and decrease in the intervals on which the graph for the function is concave up and concave down. Then sketch the curve defined by this function. Where we have Y equals X 1/3, plus X minus 2 1/3. We also have Y, Y double prime, and a graph given here. To plot our equation. Now, let's first determine Whether it's increasing or decreasing, and on what intervals. To do this, we will first Check the first derivative. Our first er that is Y equals 1/3, multiplied by 1 divided by X 2/3, plus 1 divided by X minus 2 to 2/3. Now the domain of this function overall. is going to be negative infinity to infinity, because we have cubic functions in what. If we look at our first derivative. 1/3, multiplied by 1 divide of X to 2/3. Plus 1 divided by X minus 2 to the 2/3. If we look at this function here. This value will always be positive. Except Where X equals 0, and X equals 2. We would get undefined values at those points. This then tells us that Y It's always positive. Now, we do have those undefined points. But if we ignore those undefined values, this function, is always increasing. And this is based on our first derivative. And so we'll say increasing from native infinity to infinity. Now, let's find inflection points and concavity by checking the second derivative. Now, our second derivative is -29s multiplied by 1 divided by X to the 5/3, plus 1 divided by X minus 2 to 5/3. We can simplify this to be -29s multiplied by X to negative 5/3, plus X minus 2 to the negative5/3. Now, to find the inflection point, we will set this equals to 0. We have 0 equals -29, X-5/3. Plus x minus 2 to 5/3. We can get rid of our two nights by dividing on both sides. To get 0 equals. X to the negative 5/3. Plus X minus 2 to the negative 5/3. Now let's simplify. If we shift our functions around, we end up getting the cube root of X. Equals the native Hebrew. of X minus 2. We would subtract one of our functions divided by the negative. And then simplify our exponential. No. Let's cube both sides of the equation. This will get rid of the roots. To give us X equals negative X minus 2. Or in other words, X equals negative X + 2. Or 2 X equals 2. Divide 2 on both sides to get X equals 1. That's one of our possible inflection points. But let's analyze the signs of this. So, For X is less than 0. If we plug in a value, will end up getting a negative value. So since we have a negative output, this tells us that this Is concave up. Now we have between 0 and 1. Let's let X equals 1/2, for example. The second derivative of 1/2, which is Y. is -0.59 approximately. So, because that is part, is negative there. This part ends up being concave down. No, let's check. Or between 1 and 2 Why the low prime of let's say 3 Hs. This equals to 0.59. Which makes this concave up, since it's a positive value. And then finally, X greater than 2. Let's say we plug in some arbitrary value, but we would notice. That this value will be negative. So this is also a concave down. We now have our concavity set. We also have concavity changes. At X equals 0, X equals 1, X equals 2, so these are inflection points. Now, we can go ahead and graph our function. Let's find some example points. In particular, we can find our intercepts. Will that Y equals 0. And if you look at our equation. We had 0 equals our function. 0 equals X to the 1/3. Plus X minus 2 to 1/3. We can solve this, and we end up getting X equals 1, giving us a point of 10. We can also plug in some more values, say X equals 0, and X equals 2. We end up getting Y equals. Now the 1.26. And Y equals 1.26. We can now graph all of our values. We have a 0, -1.26. 10 And 2 1.26. This is A root function, so it will open up to the right and down to the left. This function looks similar to this, where we have a curve. Inflection point At our X intercepts. Then we have 10 and another inflection point. At 2 And then we just go up further to the rights. OK, hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 49–52, graph each function. Then use the function’s first derivative to explain what you see.
y = 𝓍²/³ + (𝓍―1)²/³

Key Concepts

Graphing Functions

First Derivative

Critical Points and Behavior Analysis

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