First Derivative Test

a. Locate the critical ...

Question

First Derivative Test

a. Locate the critical points of f.

b. Use the First Derivative Test to locate the local maximum and minimum values.

c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

f(x) = x - 2 tan⁻¹ x on [-√3,√3)

Accepted Answer

Hello, in this video, we are going to be using the first derivative test and identifying the absolute extrema of the given function. The function given to us is F of X equal to X plus 2 cotangent inverse of X, and we want to go ahead and identify the following information on the defined interval from negative square root of 3 to square root of 3. Now, the first part of the problem asks us to find the critical points and use the first derivative test to classify these points. In order to find the critical points, we first need to calculate the derivative of the given function. We will calculate the derivative by taking the derivatives of the independent terms. The derivative of X will be 1. And the derivative of two cotangent inverse effects by definition, is going to be 2 multiplied by -1 divided by 1 + X squared. Next we will go ahead and simplify the derivative. We can multiply 2 by the derivative of cotangent inverse, and rewrite the derivative as 1 minus 2 divided by 1 + X2. We can then represent one as a fraction with a similar denominator by rewriting it as 1 + X2, divided by 1 + X2 minus 2 divided by 1 + X2. By combining the fractions together and simplifying like terms, F of X can be simplified to be X2 minus 1 divided by 1 + X squared. This is going to be the simplest form of the derivative. Now, what we want to do is we want to calculate the critical points of this derivative. Now, since the derivative is in the form of a rational function, we want to find the values of X that cause the numerator of the derivative to be zero. In order to find this, we will take the derivative X2 minus 1, set it equal to 0, and solve. By adding one to both sides of the equation, we get X2 equal to 1, and by taking the square root of both sides, we get X equal to plus or minus 1. What this means is that we have 2 critical points. The first critical point is at 1, and the second critical point is at -1, and this is going to be the first solution to this problem. Now we want to use the first derivative test to classify these points as local maximums or local minimums. In order to do this, we are going to create a number line. Now, on this number line, we are going to plot our boundary values. That is going to be negative square root of 3, and square root of 3. Then we will go ahead and plot the critical points. The critical points will be located at -1 and 1. Now, what we want to do is we want to take testing values for the independent intervals of the number line, plug it into the derivative, and depending on whether we have a positive or negative value, that will tell us whether the graph F of X is increasing or decreasing. So, what is the testing point that we can take between square root of 3 and -1? Well, that testing point can be negative too. If we take the value of -2, and plug it into the derivative, F of -2 will give us the value of 35. Since the derivative is positive in this region, the graph F of X will be increasing. A testing point between -1 and 1 is 0. If we plug in 0 to the derivative, F of 0 will simplify to be -1. Since the value is negative of the derivative, the original function will be decreasing within this region. And a testing point that we can take between 1 and square root of 3 is 2. F of 2 will give us the value of 3/5, and that tells us that the graph F of X will be increasing within this region. Now, by the first derivative test, since the intervals around -1 switch from increasing to decreasing, this classifies -1 to be a local maximum. And by the first derivative test, since the graph switches between decreasing to increasing, when it passes through positive 11 will be a local minimum. So again, we have a local max. Located at X is equal to -1, and we have a local minimum located at X's equal to 1. This is going to be the 2nd solution to this problem. Now that we've identified the local maximmins, let's go ahead and move on to the second part of this problem. Now what we want to do is you want to identify the absolute maximums and absolute minimums. In order to do this, we are going to take the values of our boundary points and the critical points, and plug them into the original function. Then we are going to compare the output values to each other. So, let's go ahead and take the first boundary point, negative square root of 3, and plug it in to the original function. Plugging in square root of 3 will give us the function or the value negative square root of 3, plus 2 multiplied by 5 pi divided by 6, and this will have an approximate output of 3.5. By plugging in the first critical point, F of -1 will give us the value -1 + 2 multiplied by 3 pi divided by 4, and this will have an approximate output of 3.7. Next, F1 will give us the value 1 + 2 multiplied by pi divided by 4, which has an approximate output of 2.6. And finally, by plugging in the last boundary point, F square root of 3 will give us the value square root of 3 plus 2 multiplied by pi divided by 6, which is approximately equal to 2.8. Now, let's go ahead and compare these values and start by finding the absolute. Maximum Now, when we compare the four values together, which of these values has the highest output? Well, that value is going to be 3.7, located at the X value -1. What this means is that we have an absolute maximum at the X value -1, with the output being 3. or -1 plus 3 pi divided by 2. Next, what about the absolute minimum? While between the four values, which one has the lowest output? While the lowest output between the four is going to be 2.6, at the value X is equal to 1. So, at the value X is equal to 1, we have an absolute minimum with the output being 1 plus pi divided by 2, and this is going to be the final solutions to this given problem. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

First Derivative Testa. Locate the critical points of f.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).f(x) = x - 2 tan⁻¹ x on [-√3,√3)

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First Derivative Test

a. Locate the critical points of f.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

f(x) = x - 2 tan⁻¹ x on [-√3,√3)