62–65. {Use of Tech} Graphing f and f'
c. Verify th...

Question

62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=(sec^−1 x)/x on [1,∞)

Accepted Answer

Hello. In this video, we are going to be approximating the value of X that corresponds to a horizontal tangent line. So, we are told to consider the function F of X, which is equal to seek an inverse of 2 X divided by X on the interval 1/2 to infinity. We want to approximate the value of X that corresponds to a point where F of X has a horizontal tangent line by graphing F of X using the graphing utility. So how do we solve this problem? Well, we want to approximate a value of X at a horizontal tangent line of our graph F of X. Since we are looking for a value located at a horizontal tangent line, what this means is that we need to first find the derivative of our given F of X function. Now, the problem tells us that our function F of X is equal to sequent inverse of 2 X divided by X. How do we take the derivative of this function? Well, notice that the function is given to us as a rational function. But we do have a function in the numerator that is going to require the chain rule. So, before we jump into using the quotient rule, let's quickly review what the derivative of seeking inverse of 2 X is. If we want to take the derivative of sequent inverse of X, that derivative is defined as 1 divided by X multiplied by the square root of X2 + 1. What this means is that if we take the derivative of seeking inverse of X, whatever we have for the X variable is going to be placed in all the X's for the denominator. Now, what's tricky about our seek and inverse is that we have 2 X. So we're going to have to use the definition of seek and inverse of X, but we will have to multiply by the derivative of the innermost function by chain rule. So, let's start the quotient rule and label the numerator as its independent F of X function, sequent inverse of 2 X. Now let's go ahead and take the derivative. By the definition of seekin inverse of X, since 2 x is located in the argument of skin inverse, then we will replace every X in the definition with 2 X. So the base derivative will be 1 divided by 2 X multiplied by the square root of 2 X squared plus 1. Then by the chain rule, we will need to multiply this by the derivative of the innermost function, which is 2 X, and by the power rule that derivative will be 2. This will then further simplify to give us 2 divided by 2 X multiplied by the square root of 4 X 2 because 2 X2d will simplify to 4 X 2 plus 1. This is going to be the derivative of our numerator. Now, for the quotient rule, we will label the denominator as G of X, and that is going to equal to X. And by the power rule, the derivative of the denominator will just be one. So now that we have the functions for the numerator and denominator and the respective derivatives, let's go ahead and now use the quotient rule to find the overall derivative for F of X. So F of X will equal to G of X, which is X multiplied by the derivative of F of X, which will be 2 divided by 2 X multiplied by the square root of 4 X2 + 1. Minus G of X, or I'm sorry, minus F of X, which will be sequent inverse of 2 X. Multiplied by the derivative of the denominator, G of X, which will just be 1, and seeking inverse 2 X multiplied by 1 will just be itself. Then this will all be divided by the original denominator squared, which will be X2d. Now, there's one thing that we can do to simplify the numerator. First, since we have a multiple of 2 in both the numerator and denominator, we can reduce those to be 1. And since we have an X factor in the denominator and we are multiplying by X, we can reduce those as well. This will further simplify the derivative to be 1 divided by the square root of 4 X2 + 1 minus seek an inverse of 2 X all divided by X2. Now what we need to do is we need to simplify the complex fraction. If we want to do this, why don't we go ahead and multiply the numerator and denominator by the lowest common multiple of all the denominators in our complex fraction. That lowest common multiple. is going to be X2 multiplied by the square root of 4 X2 + 1. Multiplying this throughout the denominator, but throughout the derivative will leave us with X squared. Minus sequent inverse of 2 X multiplied by X2, multiplied by the square root of 4 X2 minus 1, all divided by X to the 4th power. Finally, what we can do is we can factor out an X squared and eliminate those terms from the numerator and denominator, and the simplest form of our derivative will be 1 minus sequent inverse of 2 X multiplied by the square root of 4 X2 minus 1, all divided by X2. This is going to be the simplest form of our derivative F of X. Now that we have the derivative, we want to approximate the value of X for the original function where the horizontal asymptote exists. Now the horizontal tangent line. will exist when the derivative F of X is equal to 0. So let's go ahead and take the derivative that we just saw for, and plug it into a graphing utility and see where we have a horizontal tangent line. So what I have here is the graph of the derivative F of X that we just solved for. Now where would the horizontal tangent line exist for the graph of the derivative? While the horizontal tangent line would exist at the X intercept of this graph, and that X intercept exists at the point 0.76665.0. And if we were to approximate the X value to two decimal places, X would approximately equal to 0.77. That means that the approximate value of X for the original function where the horizontal tangent line exists, is going to approximately be 0.77. And with that being said, the overall solution is going to be D. 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.

62–65. {Use of Tech} Graphing f and f'
c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.
f(x)=(sec^−1 x)/x on [1,∞)

Key Concepts

Derivative and Critical Points

Horizontal Tangent Lines

Graphing Functions and Their Derivatives

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