Slopes and Tangent Lines

In Exercises 13–16, ...

Question

Slopes and Tangent Lines

In Exercises 13–16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable.

f(x) = x + 9/x, x = −3

Accepted Answer

Below there today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Using the limit definition, find the derivative of the function and the slope of the tangent line at the given value of X. F of X is equal to X plus 15 divided by X, X is equal to -5. Awesome. So it appears for this particular problem we're asked to use the limit definition to help us find the derivative of the function and the slope of the tangent line at the given value of X. Awesome. So it appears for this particular problem we're trying to solve for two separate answers ultimately. Our first answer is we're trying to find this derivative of this function of F of X. That's our first answer, and our second answer is we're trying to determine the slope of the tangent line at the given value of X. Awesome. So now that we know that we're trying to solve for two separate answers, our first step to solve for this particular problem is we need to recall and use or recall and apply the limit definition of the derivative. So, as we should recall, the derivative is defined as F of X, where this prime symbol means the first derivative of F of X. So F of X is equal to the limit, as H approaches 0 of F of X plus H minus F of X all divided by H, fantastic. So now our next step. As we need to note that F of X, we need to note that F of X is equal to X plus 15 divided by X, and we need to note that F of X plus H is going to be equal to X plus H plus. 15 divided by X plus H, we can go ahead and write that F of X is going to be equal to the limit as H approaches 0 of X plus H. Plus 15 divided by X plus H minus parentheses X plus 15 divided by X. All divided by H. Fantastic. And when we do a little bit of simplifying, we will find that F of X is going to be equal to the limit, as H approaches 0 of H plus 15 divided by X plus H. Minus 15 divided by X, and then this will be all divided by H. Fantastic. So our second step now is we need to simplify our expression, so we need to take F of X is equal to the limit as H approaches 0 of square bracket. H divided by H plus 1 divided by H multiplied by parentheses, 15 divided by X plus H minus 15 divided by X, and don't forget your square bracket on the end. Moving right along, so then we can write that F of X, when we do a little bit simplifying, is equal to 1 plus the limit as H approaches 0 of 1 divided by H multiplied by parenthesis. 15 X minus 15 multiplied by X plus H in parentheses, all divided by X multiplied by parentheses X plus H. Fantastic. So, moving right along, we do a little bit of simplifying and we'll find that F of X is gonna be equal to 1 plus the limit as H approaches 0 of 15 X. Minus 15 X minus 15H. Fantastic. All divided by H X or H multiplied by X, multiplied by parentheses X plus H. Fantastic. Moving right along here. So, we can go ahead and write that F of X, when we do a little bit of more simplifying, is going to be equal to 1 plus the limit as H approaches 0 of negative 15 H divided by HX multiplied by parentheses X plus H, fantastic. And now, We can simplify one last time, we could say that F of X is equal to 1 plus the limit as H approaches 0 of -15 divided by X multiplied by parentheses X plus H. Fantastic. So now, our next step. we need to evaluate the limits. So that's our 3rd step. So, step number 3 is we're trying to evaluate the limit, so we need to note that as H approaches 0, that will mean that X plus H will approach X. So that means we can go ahead and write that F of X is going to be equal to 1. Plus 15 divided by X multiplied by X. So that will mean that F of X is going to be equal to 1 minus 15 divided by X to the power of 2. So thus, the derivative of F of X is equal to X plus 15 divided by X will be F of X is equal to 1 minus 15 divided by X2. So that's our first answer that we're ultimately trying to solve for. Hooray, we did it. So now we can start solving for our second answer. So our fourth step is we need to find the slope of the tangent line at X is equal to -5. So we need to recall and note that the slope of the tangent, or rather the slope of the line tangent to the curve at X equals -5 will be F of -5. So we need to substitute X equals -5 into our derivative, so we need to say that F of -5 is going to be equal to 1 minus. 15 divided by -5 to the power of 2. So when we evaluate that expression, which I'm going to skip a few steps, but essentially when we perform that calculation or calculator, so 1 minus 15 divided by -5 to 2, you'll find that F of -5, when we write it in its most simple fraction form, will be equal to 25 or 2 divided by 5. And that's it. That is our final answer for the slope of the tangent line. Hooray, we did it. So, going back to look at our problem and to quickly recap everything that we've learned, our final two answers, without a doubt, have to be. F of X is equal to 1 minus 15 divided by X2, and F of -5 is equal to 25 or 2 divided by 5, and those are our final two answers for both the derivative of our function F of X, which is our first answer, and then the slope of the tangent line at the given value of X, which the given value of X once again was X equals -5. And that's it. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Slopes and Tangent Lines

In Exercises 13–16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable.

f(x) = x + 9/x, x = −3

Key Concepts

Differentiation

Slope of the Tangent Line

Evaluating Derivatives at a Point

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