Increasing and decreasing functions. Find the intervals on whi...

Question

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.

f(x) = x² - 2 ln x

Accepted Answer

Hello. In this video, we are going to be identifying the intervals of increasing and decreasing. We are asked to determine the intervals on which the function G of X equal to X cubed minus 3 natural logarithm of X is increasing or decreasing. Now, in order to identify or find these intervals, the first thing we need to do is calculate the critical points of the function. In order to calculate the critical points, we first need to find the derivative. The function given to us is G X equal to X cubed minus 3 multiplied by the natural logarithm of X. We will go ahead and take the derivative of this function with respect to X. In order to take the derivative, we will take to the derivative of each term. The derivative of X cubed will be 3 x squared. And the derivative of -3 natural logarithm of X will be -3, multiplied by the derivative of the natural logarithm of X, which is 1 divided by X. Next, what we will go ahead and do is simplify the derivative. We can rewrite the derivative to be 3 X2 minus 3 divided by X. If we combine this into a single fraction, we can rewrite 3 X squared as 3 x cubed divided by X minus 3 divided by X. And if we combine these two fractions into a single fraction, the derivative G of X will simplify to be 3 x cubed minus 3 divided by X. So Let's go ahead and first determine the domain of the original function. Now, the first portion of the function is all real numbers, since it is a polynomial, but since we have a natural logarithm of X within our function, the domain is going to be all values of X that are strictly greater than 0. So, what this means is that for our derivative, we want to work with values of X that are strictly greater than 0. To calculate the critical points, we will take the values of X in the numerator that caused the numerator to become 0. So we will take our numerator, 3 X cubed minus 3, set it equal to 0, and solve. We can add 3 to both sides of the equation, giving us 3 X cubed equal to 3. We will divide both sides of the equation by 3, giving us X cubed equal to 1, and by taking the cube root of both sides of the equation, we get X is equal to 1. Now, since 1 is obtained in our in our domain, X greater than 0, that means that we only have the one critical point located at X is equal to 1. So, now what we are going to do is we are going to create a number line. This number line is going to start at the value of 0 and go off to positive infinity because of the domain of the natural logarithm. Then we will go ahead and plot the value of our critical point, which is 1. What we want to do now is you want to take testing points from the interval from 0 to 1 and from 1 to positive infinity. We will take these testing points and plug it into our derivative. If the derivative is positive, then the graph G of X is increasing in that region, and if the derivative is negative, then the graph G of X will be decreasing in that region. What is a testing point that we can take between 0 and 1? A testing point that we can use is 0.5. If we plug in 0.5 into G of X, G of 0.5, will give us the value of 3 multiplied by 0.5 cubed minus 3, all divided by 0.5, and this will give us the approximate value of negative 5.25. Since this value was negative, this tells us that the graph G of X will be decreasing on the interval from 0 to 1. We will repeat this process for the interval from one to infinity. A testing point we can take from 1 to infinity is 2. Plugging this into the derivative will give us G 2 equal to 3 multiplied by 2 cubed minus 3. And this will all be divided by 2, which will simplify to give us 21 divided by 2. Since this value is positive, that tells us that the derivative will be or the original function will be increasing on the interval from one to infinity. So to summarize, The interval of increasing of G of X is going to be the interval from 1 to positive infinity. And the interval of decreasing for the original function is going to be the interval from 0 to 1. This is going to be the solution to the problem. And the overall solution is going to be option A. So I hope this video helps you in understanding on how to find the intervals of increasing and decreasing, and I will go ahead and see you all in the next video.

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.

f(x) = x² - 2 ln x

Key Concepts

Derivative

Critical Points

Test Intervals

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