Theory and Examples

In Exerc...

Question

Theory and Examples

In Exercises 51 and 52, give reasons for your answers.

Let f(x) = (x − 2)²ᐟ³.

b. Show that the only local extreme value of f occurs at x = 2.

Accepted Answer

Welcome back, everyone. Let G of X be equal to x minus 4, raise the power of 2/5. Which of the following statements is true about the local extreme values of G of X? A says G of X has a local maximum at X equals 4. B G of X has a local minimum at X equals 4. C G of X has both a local minimum and a local maximum at x equals 4, and D G of X has no local extreme values. If we want to identify the extreme values of a function, we first will have to differentiate it. So we're going to identify G of X. Which is the derivative of X minus 4 raised to the power of 2/5. We are applying the chain rule. First of all, by applying the power rule, we're going to get 25 multiplied by X minus 4 raise to the power of 25 minus 1, which is -35, and according to the chain rule, we're multiplying by the derivative of X minus 4, which is simply equal to 1. So we can neglect it, right? Meaning our first derivative is 2. Divided by 5, multiplied by X minus 4, raises the power of 3/5. So for simplicity here, we are applying the reciprocal rule for fractions. Now if we want to understand whether or not we have local minimum or local maximum at X equals 4, well, we essentially have to understand that we're looking for the critical points, right? So we have two cases whenever we're constraining critical points. The first case is to set our derivative equals to 0. And we can say that there are no solutions, right, because we have a fractional expression and our numerator is independent of X. Now, the second case is to check where the first derivative does not exist. And we want to ensure that. The critical point that we identify is within the domain. For a rational function, this means that X equals 4 because we get a denominator of 0, right, and division by 0 is undefined. So that's our critical point. And it is also in the domain of the original function x minus 4. Raise the power of 2/5 was the original function. We can rewrite it as fifth root of X minus 4 squad. So essentially we're taking 5th root of a number, and besides our number corresponds to a square, right, so we never get a negative number. In every case, for every X, our function is defined. So now that we have our critical point, which is in the domain of the original function because the domain is all real numbers, we want to classify it, and what we can do is simply perform the first derivative test. Essentially, If we consider G at any value which is less than 4, such as G at 3. It is going to be equal to 2, divided by 5 multiplied by 3 minus 1, raise the power of 3/5. We have to understand that. This expression, I'm sorry, it should be 2 divided by 5 multiplied by 3 minus 4 races to the power of 3/5, right? So this expression is going to be negative because we have a positive number divided by a positive number and again divided by a negative number, 3 minus 4 is -1, and when we erase a negative number to an exponent, we get a negative number. And now if we evaluate G at any point that is greater than the critical point, such as G at 5, we can show that this is 2 divided by 5 multiplied by. 5 minus 4 raise to the power of 3/5. So in this case we are raising 1 to the power of 3/5, which is a positive number. So the overall. Ratio would be positive. Now what does that mean? Well, if we plot a number line with the critical point of 4. We now know that for any X that is less than 4, the derivative is negative. And for any X that is greater than 4, the derivative is positive. So a negative derivative represents that our function is decreasing. And then, after the critical point of 4, our function starts to increase. So X equals 4 is a local minimum. Therefore, we can conclude that B is the correct answer. G of X has a local minimum at x equals 4. Thank you for watching.

Theory and Examples

In Exercises 51 and 52, give reasons for your answers.

Let f(x) = (x − 2)²ᐟ³.

b. Show that the only local extreme value of f occurs at x = 2.

Key Concepts

Local Extreme Values

First Derivative Test

Critical Points

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