Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x¹ᐟ³(x + 8)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to analyze the function H of X, which is equal to X-rays to the 1/3 power, multiplied by the quantity of X minus 4 in quantity to determine where it is increasing or decreasing and to identify any local extreme values. Now, to solve this problem we first need to find are critical points. And so for that we need to calculate H of X. And H X is going to be equal to the derivative with respect to X. Of quantity of X-rays to the 1/3 power multiplied by the quantity of X minus 4 in quantity. And here we're taking a derivative of a product, so we need to use our product rule. So recall for the product rule that if you have a function Y that is equal to F of X multiplied by G of X. Then its derivative Y is equal to F multiplied by G plus F multiplied by G. So in our case, F is going to be equal to X-rays to the 1/3 power and G is going to be equal to x minus 4. So F is going to be equal to the derivative with respect to X. Of X-rays to the 1/3 power. And here we can apply our power rules to take this derivative, and that's going to be equal to 1 divided by 3, multiplied by X-rays to the minus 2/3 power. And for G, that's going to be equal to the derivative with respect to X of the quantity of X minus 4. And again, using our power rule as well as our concept rule for a derivative, this is going to be equal to one. And so now we can actually calculate our derivative HX using our product rule. So H X is going to be equal to F, which is 1 divided by 3, multiplied by X raised to the minus 2/3 power, and then that gets multiplied by G, which is the quantity of X minus 4 in quantity. Then we'll have plus F which is X raised to the 1/3 power multiplied by G which is 1. So what we're going to do now is combine both of these two terms with a common denominator. So our common denominator is going to be 3 multiplied by X-rays to the 2/3 power. So, combining these two terms, this is going to be equal to the quantity of X minus 4. Plus 3 x in quantity divided by The quantity of 3 multiplied by x raised to the 2/3 power in quantity. Combining like terms in the numerator, this is going to be equal to the quantity of 4 x minus 4 in quantity divided by the quantity of 3 multiplied by x raised to the 2/3 power in quantity. So now we need to find our critical points and recall that our critical points occur whenever our first derivative is either undefined or equal to zero. So our first derivative is undefined in this case whenever our denominator is equal to 0, and that occurs when X is equal to 0. So that is one of our critical points. The other critical point is going to come from our first derivative being equal to 0, and that's going to occur whenever our numerator is equal to 0. So that means we'll have 4 X minus 4 equal to 0, and solving this for X, we get X is equal to 1. So we're gonna use these two critical points to determine the intervals over which our function is increasing or decreasing. So the first interval that we're going to look at is the open interval from minus infinity to 0. And here we need to determine the sign of our first derivative, since the sign of the first derivative will tell us whether our function is increasing or decreasing. And the sign of our first derivative H X is determined entirely by the sign of the numerator, since our denominator is always going to be a positive quantity. So here I'm just going to look at the quantity of 4 X minus 4. And when X is less than 0, this quantity is also less than 0. And since that's less than 0, that means that H of X. It's also less than 0. And since H X is negative, recall that if the first derivative is negative over an interval, that means our function H of X is decreasing. So our function is decreasing over this interval. The next interval that we're going to look at is going to be the open interval from 0 to 1. And again we're going to focus on just the numerator, so the quantity of 4 x minus 4, and on this interval, the quantity of 4 X is still going to be less than 4. And so when I subtract 4 from this, that means that this quantity 4 X minus 4 is going to be less than 0. And so that means that H of X is also going to be less than 0. And again, since H X is negative, that means our function H of X is decreasing on this interval. The last interval that we want to look at is going to be the open interval from one to infinity. And again, focusing on the quantity of 4 X minus 4. When X is greater than 14 X is actually going to be greater than 4, so 4 x minus 4 is also going to be greater than 0, and that means that H of X is going to be greater than 0. And since our first derivative is positive on this interval, that means our function H of X is increasing on this interval. So based on this information, we see that the critical point at X equal to 0 is not a local extrema. Since our function is decreasing on the intervals both to the left and right of X equal to 0, but we do have a local minimum at X equal to 1, and that's because H of X is changing from decreasing to increasing. So now we need to find the value of H X when X is equal to 1. So we'll calculate H of 1, and that's going to be equal to 1 raised to the 1/3 power multiplied by the quantity of 1 minus 4 in quantity, and if I win this expression, this is going to be equal to -3. So we now have all the information that we need to write our final answer. So if we look at the intervals over which our function is increasing or decreasing, we see that our function is increasing. Over the open interval from one to infinity. And if we look at the intervals over which our function is decreasing, We see that it is decreasing on the open interval from minus infinity to 1. And we actually have a local. Minimum When X is equal to 1. And we have H of 1 equal to minus 3. And if we look for local maxima, We see that we actually have none. So these are the answers that this problem was looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x¹ᐟ³(x + 8)

Key Concepts

Derivative and Critical Points

First Derivative Test

Interval Analysis

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