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

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)=e^−x tan^−1 x on [0,∞)

Accepted Answer

Hello. In this video, we are going to be working on the following tangent line problem. So we are told to consider the function F of X is equal to 3 multiplied by E raises the power of negative X multiplied by tangent inverse of 3 X on the interval from 0 to infinity. We want to approximate the value of X that corresponds to a point where F has a horizontal tangent line by graphing the derivative F using a graphing utility. So how do we approach this problem? While we are asked to approximate a value of X around a horizontal tangent line. Since we are focused on using a horizontal tangent line that indicates that we are going to need the derivative of the original function F of X. So, the function given to us is F of X is equal to 3 multiplied by E raised the power of negative X multiplied by tangent inverse of 3 X. So, how do we take the derivative of this function? Well, notice that our function is given to us in the form of the product of two functions. So in order to take the derivative, we are going to need to use the product rule. In order to take the product rule, let's go ahead and first take the derivative of our first function F of X, which will put 3 multiplied by Ease the power of negative X. In order to take the derivative of F of X, we are going to need to use the exponential rule. So that is going to be. 3, multiplied by the derivative of E rates to the negative X. We will take the original function E raised to the negative X, then multiply that by the natural logarithm of the base of the function, which is E, then multiply this by the derivative of the exponential function, which is going to be negative X. Now let's go ahead and simplify. The natural logarithm of E is just going to be 1. And the derivative of negative X by the power rule is going to simplify to be -1. So, this is going to leave us with 3 multiplied by negative Ease the power of negative X, which will further simplify to be -3, multiplied by Ease the power of negative X. This is going to be the derivative of the first function. Now let's take the derivative of the 2nd function. So, let's quickly recall what the derivative of tangent inverse is. The derivative of tangent inverse of X. is given to us as one divided by 1 plus X squared. This means that whatever argument we have for X, that argument is going to fill in every X in the definition. But since we have a function of X within this tangent inverse function, we are going to need to use the chain rule. So, in this case, our X is going to be 3 X, and the derivative of tangent inverse of 3 X is going to be 1 divided by 1 plus the quantity 3X squared, then by the chain rule we must multiply this by the derivative of the innermost function 3X. That derivative is going to be 3. Further simplifying the derivative, this is going to give us 3 divided by 1 plus 9 X squared. This is going to be the derivative of the second function, which we will label as G of X. Now that we have the derivatives of the 1st and 2nd function, and we know the original 1st and 2nd function, we can now use the product rule to simplify the derivative. So, that is going to be F of X is equal to the original first function, which is 3, multiplied by E raise the power of negative X, multiplied by the derivative of the second function tangent inverse of 3 X. That derivative is going to be 3 divided by 1 + 9 X squared. Plus, the original second function, which will be tangent inverse of 3 X, multiplied by the derivative of the first function, which will be -3, multiplied by E, raise to the power of negative X. Now let's go ahead and simplify. So, multiplying -3, multiplied by E raise to negative X with tangent inverse is going to leave us with 3 multiplied by Ease the power of negative X. Multiplied by 3 divided by 1 + 9 X squared. Minus 3 tangent inverse of 3 X. Multiplied by E raised to the power of negative X. Now notice that both of the terms in the difference have a factor of 3 multiplied by Ease the power of negative X. If we factor out these terms, that is going to give us 3 E to the negative X multiplied by 3 divided. By 1 + 9 X2 minus tangent inverse of 3 X. This is going to be the simplest form of our derivative. Now, what we want to do is we want to calculate or approximate the value of X by using the horizontal tangent line. Now, recall that the horizontal tangent line occurs when the derivative F of X is equal to 0. So, let's go ahead and use a graphing utility to figure out where F of X is equal to 0. So what I have here is the graph of our derivative. Now, this is the graph of F of X. The horizontal tanguline occurs when this derivative is equal to zero, which means that the horizontal tanguline will occur at the X intercept of the graph. That X intercept will be located here at the point. 0.48331.0. This means that the approximate value of X, when approximated to two decimal points of the original function, is going to be 0.48, and this is going to be the solution to this problem. And with that being said, the overall solution is going to be B. 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)=e^−x tan^−1 x on [0,∞)

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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)=e^−x tan^−1 x on [0,∞)