Each of Exercises 67–88 gives the first derivative of a contin...

Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
71. y' = x(x² - 12)

71. y' = x(x² - 12)

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says the first derivative of a continuous function H of X is H X, which is equal to X multiplied by the quantity of X2 minus 27 in quantity. Find the second derivative of H of X and sketch the general shape of its graph. Below the problem we're given an area over which to make the sketch of our graph. Now the first part of this problem wants us to find the second derivative. And we were given H X. So to find the second derivative, which is H of X, we just need to take a derivative with respect to X of H X, which is going to be the quantity, and here we'll multiply out our function H X. So that will give us X cubed minus 27 X in quantity. So taking this derivative, using our power rule and constant multiple rule, this is going to be equal to 3 x 2. -27. And so that is the answer for the first half of this problem. Now, for the second half, we need to sketch the general shape of a graph. And so the first thing we need to do is find local extrema, and for that we need to find critical points. And recall that critical points are located whenever our first derivative is equal to zero or undefined. Now, H of X is a polynomial, so it's defined everywhere. So the only time we're going to have critical points is when H of X is equal to 0. So we'll set X multiplied by the quantity of X2 minus 27 in quantity equal to 0. So we'll set each factor here equal to 0 and solve for X. So for the first factor, we'll just get X equal to 0, and for the second factor, we'll have X2 minus 27 is equal to 0. So solving for X by adding 27 to both sides, we'll get x 2 is equal to 27, and then taking the square root of both sides, we'll get X. It's equal to plus or minus 3 multiplied by the square root of 3. And so now we're going to use our critical points to determine the intervals over which our function is increasing or decreasing. So for this, we'll create a table. In the first row we'll have the sign of H X, and in the second row, we'll have the behavior of H of X, whether it's increasing or decreasing. And the first interval that we're going to look at is going to be going from minus infinity to -3 multiplied by the square root of 3. Now on this interval, Our X value is negative, so the first factor in H X is going to be negative. But, if we look at the second factor here, The quantity of X2 is actually going to be always greater than 27. Over this interval. So when I subtract 27, X2 minus 27, it's going to be a positive quantity. So we'll have a negative value multiplied by a positive value, which will give me an overall negative quantity for H X. And recall that if H X is negative, that means that our function H of X is going to be decreasing. So we'll label that with a D in the second row of our table. Now, the next interval that we want to look at is going to go from -3 multiplied by the square 32 X equal to 0. Now, over this interval, X is still negative, but X2d is actually going to be less than 27. And so, X2 minus 27 is going to be a negative quantity. So we'll have a negative value multiplied by a negative value, which will give me an overall positive sign for H X. So we'll place a plus sign in the first row of the table. And recall that if the first derivative is positive, that means that H of X is going to be increasing over the same unfold. So we'll label that with an I. Now, the next interval that we want to look at is going from X equal to 0 to X equal to 3 multiplied by the square root of 3. And over this interval, X is positive, so our first factor in H X is positive, but the quantity of X2d is always going to be less than 27, so the factor of X2 minus 27 is going to be negative. So I'll have a positive value multiplied by negative value, which will give me an overall minus sign for H X, and that means that our function H of X is decreasing over this interval. And then the final interval that we want to look at is going from X equal to 3 multiplied by square root 3 to infinity. And over this interval, we see that X is going to be positive, and the quantity of X2 is going to be greater than 27, so X2 minus 27 is also going to be positive. So that means I'll have the multiplication of two positive quantities. So H X is going to be positive, and that means that our function is going to be increasing over this interval. So we see that we have local minima at X equal to plus or minus 3 multiplied by the square root 3, since our function H of X is changing from decreasing to increasing at those points, and we'll have a local maximum at X equal to 0, since our function is changing from increasing to decreasing at that point. Now, the next thing we need to do is determine our inflection points, and recall that inflection points occur whenever our second derivative is equal to 0. So we're going to set 3 X2 minus 27 equal to 0 and solve for X. So, the first thing we're going to do is add 27 to both sides, and so we'll have 3 X squared, it's equal to 27. Then we'll divide both sides by 3, so we'll get X2d is equal to 9, and then taking the square root of both sides, we'll get X is equal to plus or minus 3. So now we need to determine whether these Two values are actually inflection points, and for this we'll need to determine our intervals of concavity. So again, we'll do this with a table, where the first row is going to be the sign of each double prime of X. And the second row is going to be the behavior of HF X, whether it's concave up or concave down. So the first interval that we'll look at, is going from minus infinity to X equal to -3. Now, over this interval, X2d is going to be greater than 9, and so 3 multiplied by a quantity that's greater than 9, is going to be greater than 27. And when I subtract 27 from that, I'm going to get an overall positive quantity. So that means that H X is positive over this interval, so we'll place a plus sign in the first row of our table, and recall that if the second derivative is positive over an interval, H of X is going to be concave up. So we'll label that with a U. Now, the next interval that we're going to look at is going from X equal to -3 to X equal to 3. Now, over this interval, X2 is going to be less than 9, so 3 X2 is going to be less than 27. And so when I subtract 27 from that quantity, I'm going to get a negative value. So that means that H of X is going to be negative over the interval. And recall that the second derivative is negative, that means that our function H of X is going to be concave down. So we'll live that with a D in the second row of our table. And the final interval that we want to look at is going from X equal to 3 to infinity. And over this interval, X2 is going to be greater than 9, and so 3 X2 is going to be greater than 27. And when I subtract 27 from that, I'm gonna get an overall positive quantity. So, we'll place a plus sign in the first row of our table. And here, since the second derivative is positive, that means that H of X is going to be concave up. So we'll place a U in the second row of our table. So we see that we do indeed have inflection points at X equal to + -3, since the concavity is changing at those points. So we now have enough information to sketch the general shape of our graph. So we're gonna start off by placing our three critical points. So 1 at X equal to 0. 1 at X equal to. Minus 3 multiplied by the square root of 3, and another one placed at X equal to 3 multiplied by the square root of 3. And so we placed the one with X equals 0 above the other two points, since the X equals 0 correspond to a local maximum and the other 2 points correspond to a local minima. And in between these two, we're going to have two of our inflection points. So we'll have a point at X equal to minus 3 for 1 inflection point, and we'll have another inflection point at X equal to 3. And so now we just need to use our concavity and our intervals of increasing and decreasing for our function to determine um how to draw this function. So, initially our function H of X. On the interval, going from minus infinity to -3 multiplied by the squared 3, our function is going to be decreasing, and our function is also going to be concave up on this interval between minus infinity to X equal to -3. So, we'll draw our function decreasing. Until it reaches the local minimum, in which case it begins to increase. And it reaches our inflection point, and so we're changing now from concave up to concave down, but our function is still increasing, until it reaches the local maximum at X equal to 0. And then our function will be decreasing until it reaches the next inflection point at X equal to 3, in which case we're switching from concave down to concave up, and our function is going to continue to decrease until the next local minimum at X equal to 3 multiplied by the square root 3, and after which our function is going to increase until it reaches the edge of our drawing area. And so the graph that we have sketched. On the screen here is the second half to the answer to this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
71. y' = x(x² - 12)

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

Watch next

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.71. y' = x(x² - 12)

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

Watch next

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
71. y' = x(x² - 12)