In Exercises 11–18, find the slope of the function’s graph at ...

Question

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

g(x) = 8 / x², (2, 2)

Accepted Answer

Below there, today we're gonna 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. For the given function, determine the slope at the specified point and write the equation of the tangent line at that point. F of X is equal to 4 divided by X2 1.4. Awesome. So it appears for this particular problem we're asked to take our given function of F of X and we're asked to determine two separate answers. Our first answers were asked to determine the slope at this specified point. That's our first answer, and then our second answer is we're asked to write the equation of the tangent line at this specific point. So now that we know that we're ultimately trying to figure out the slope and the tangent line of this particular function, Let us first off note once again that we're told that our given function is F of X. So let's write down our given function. So F of X is equal to or divided by X squared. And let us told that we're told that it's at the point, which I'll denote point as capital P, so it's at the parentheses 1.4. And we need to note once again our goal is ultimately to determine the slope at this point using the definition of the derivative. So, as you should recall a note, the definition of the derivative states that The limit as H approaches 0 of F of X0 plus H minus F of X0 all divided by H. And let us know in this particular case that X0 will be equal to 1. So now, our first major step towards solving for our slope, we need to compute or determine. We need to determine F of X0 plus H, and we also need to determine F of X0. So let us note that F of 1 means we're plugging in a value of 1 in for x0, because that's what we determined. So when we plug in a value of 1 in for X0, you will find that F of 1 is equal to 4 divided by 1 to the power of 2, which is going to be simply equal to 4. And if we plug in F. So for F of and we're plugging in 1 for X0 plus H, so it's going to be 1 plus H, it's going to be equal to 4 divided by 1 + H to the power of 2. Fantastic. So now, our second step is we now need to compute the difference quotient, so we need to take the limit as H approaches 0 of. 4 divided by 1 plus H2 minus 4 divided by H. Fantastic. So now we need to rewrite 4 as 4 divided by 1. So when we do that, we will find that the limit as H approaches 0 of 4 divided by 1 plus H squared. Minus 4 divided by 1 squad, all divided by H. So now, if we factor out of 4, we will find that 4 multiplied by the limit as H approaches 0 of 1 divided by 1 plus H quad minus 1 divided by H. So now, if we recall in using our identity property, we will get 1 divided by 1 plus H quad minus 1 is equal to 1 minus 1 + H2, all divided by 1 plus H2. So now, when we expand this statement, we will get 1 minus 1 + 2 H + H2, which is equal to -2H minus H2. Which will be equal to when we do a little bit of simplifying, -2 H minus H2 all divided by 1 plus H2. So, therefore, we can go ahead and write that 4 multiplied by the limit as H approaches 0 of -2 H minus H2, all divided by H multiplied by 1 plus H2. So now, We need to cancel out H, and when we cancel out H, it will be equal to 4 multiplied by the limit as H approaches 0 of -2, and then we need to take -2 minus H divided by 1 + H2. So now we need to substitute in the value H equals 0. So when we substitute in the value that H equals 0, we will be able to write that 4 multiplied by -2 divided by 1 is going to be equal to -8, and this will be our value for the slope. Hooray, we did it. So M denotes the slope. So thus our slope is going to be a value of -8. Fantastic. So now we need to find the tangent line equation. So, that's our 3rd step. Our 3rd step is to find the tangent line equation. So we need to, in order to do that, we need to recall and use the point slope formula, which, as we should recall, the point slope formula states Y minus Y1 is equal to M the slope, multiplied by X minus X1. Fantastic. So we need to note in this case that M is equal to -8, and we also need to note that That X1 is going to be equal to 1, and we also need to recall that Y1 is going to be equal to 4. So Y1 is equal to 4. So when we plug in our known variables, we will get that Y minus 4 is equal to -8, multiplied by X minus 1. Fantastic. So let me rearrange this equation using a little bit of algebra to isolate and solve for Y, all by itself, you will find that Y is equal to -8 X plus 12, and that will be our final answer for the tangent line. Hooray, we did it. So going back to look at the problem itself. And to quickly recap what we've learned together, our final pair of answers have to be, without a doubt, our slope will be M equals -8, and our tangent line will be Y equals -8 X plus 12. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

g(x) = 8 / x², (2, 2)

Key Concepts

Derivative

Power Rule

Equation of a Tangent Line

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