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' = 𝓍² ― 𝓍―6

y' = 𝓍² ― 𝓍―6

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says consider the function G of X, whose first derivative is given by G of X is equal to X2 minus 2X minus 15. Termine the X coordinate at which G of X, if any, has a local maximum, local minimum, or inflection point. Now we're asking this problem to determine the X values for the local maximum, local minimum, or inflection point for a function G of X. And so the first thing we need to do is determine our critical points. Recall that our critical points are located at x values, at which our first derivative is either undefined or equal to zero. Now, our first derivative here, G X is a polynomial, and polynomials are defined everywhere. So the only option to find a critical point is to set our first derivative here equal to 0. In doing so, that means that we're going to have X2d. Minus 2 X -15 Equal to 0. And we can actually factor the quadratic on the left hand side. So we'll have the quantity of X + 3 in quantity multiplied by the quantity of X minus 5 in quantity is equal to 0. So now what we're going to do is set each one of these factors equal to 0 and solve for X. So we'll have X + 3 is equal to 0, therefore, X is equal to -3, and then we'll have X minus 5 is equal to 0, and that means that X is equal to 5. So these are our two critical points. Now, in order to determine whether these critical points are local maxima, minima, or neither, we're going to use the second derivative test. That means I'm gonna need to calculate G of X. Which is going to be equal to the derivative with respect to X of our first derivative, which is the quantity of X2 minus 2X minus 15 in quantity. So this is going to be equal to, and here to take this derivative, I'm going to use our constant multiple, our power rule, as well as our constant rule. So taking this derivative, we're going to get 2 X minus 2. So now we need to do is determine the sign of our second derivative at each of our critical points. So we're gonna start off with X equals to minus 3, so we'll calculate G of -3. And that's going to be equal to 2, multiplied by minus 3. -2, that's gonna be equal to -8. Which is less than 0, so it's negative. So by the second derivative test, that means that we have a local maximum. At X equal to minus 3. So that's gonna be our first point. Determined So that's one part of our answer. So, the next point that we need to look at is X equal to 5, so we'll calculate G of 5. And that's going to be equal to 2, multiplied by 5 minus 2, which is equal to 8, which is a positive quantity, since it's greater than 0. That means that by second derivative test, I have a local minimum at this point. So I have a local minimum. At X equal to 5. So that is another part of our question answered. So we determined the local maximum and local minimums for our two critical points. So now we need to look at inflection points. So recall that possible inflection points occur when our second derivative is equal to 0. So we're going to set 2X minus 2 equal to 0. So to solve this for X, we'll add 2 to both sides of the equation. So we'll have 2 X is equal to 2. And divide both sides by 2, we'll get X is equal to 1. And so now I need to determine if I do have an inflection point at X equal to 1. So we need to determine whether the concavity is changing for a function G of X. So what we're going to do is look at the sign of our second derivative to points just to the left and right of X equal to 1. So we'll look at our second derivative with X equal to 0, so we'll calculate G of 0, which is going to be a point to the left of X equal to 1. And this is going to be equal to 2 multiplied by 0 minus 2, which is equal to -2, and that is less than 0 since it's a negative quantity. The other point that we're going to choose to look at is going to be X equal to 2, which is a point to the right of X equal to 1. So we'll calculate G of 2, which is going to be equal to 2 multiplied by 2 minus 2, which is equal to 2, and so that is a positive quantity, so it's greater than 0. So since my sign here is changing, For a second derivative, that means the concavity of our function GFX is also changing. So that means we do have an inflection point. At X equal to 1. So we have an inflection point. At X equal to. One. So that is the last piece of our answer for this problem. So just to recap, we have a local maximum at X equal to -3. We have a local minimum at X equal to 5, and we have an inflection point at X equal to 1. So that is the answer for this problem. So I hope that this explanation has been useful, and I'll see you 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' = 𝓍² ― 𝓍―6

Key Concepts

First Derivative Test

Critical Points

Inflection Points

Watch next

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' = 𝓍² ― 𝓍―6

Key Concepts

First Derivative Test

Critical Points

Inflection Points

Watch next

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' = 𝓍² ― 𝓍―6