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.

f(x) = √(x + 1), (8, 3)

Accepted Answer

Hello 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. 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 the square root of 2 X plus 7 9.5. Awesome. So it appears for this particular problem we're asked to take our given function, and we're asked to solve for two separate answers. Our first answer is we're asked to determine the slope at the specified point, and our second answer is we're asked to write the equation of the tangent line at that specific point. So now that we know that we're ultimately trying to figure out. The slope and the tangent line of the specific function of F of X. Our first step is to examine our function that we are given. So we're told that our function is F of X is equal to the square root of 2 X plus 7. We're also told that our point, which we'll call capital P, is parentheses 9.5. Fantastic. So once again, we need to note that we're ultimately trying to determine the slope, and as we should recall, in order to determine the slope, we need to recall and use the difference quotient method. And as we should recall, we can state that the limit as H approaches 0 of F of X0 plus H minus. F of X0, all divided by H, and that is our difference quotient method expression, and we need to know in this particular case that X0 is going to be equal to 9. So now our next step, or rather our first major step towards solving for the slope is we need to compute the values of F of X0 plus H and F of X0. So F of X0. So now we need to compute these values, but let's start off first by solving for F of 9. So we're plugging in a value of 9 in for X0. So with that said, so F of 9 is going to be equal to the square root of 2 multiplied by 9 + 7. So when we plug that into our calculator, we will get an answer of 5, and for F of 9 plus H. Once again, we're plugging in a value of 9 in for X0. So F of 9 plus H is going to be equal to the square root of 2, multiplied by 9 plus H + 7. Fantastic. And this will be Equal to when we do a little bit of simplifying the square root of, and I'm gonna skip a few steps, but when we simplify, this will be equal to 25 + 2 H. Fantastic. So, moving right along here. Our second step now is we need to compute the difference quotient, so we need to take the limit as H approaches 0 of the square root of 25. Plus 2 H minus 5, all divided by H, and we need to multiply by the conjugate to simplify our answer, so we need to take Our current expression and we need to multiply it by the square root of 25 + 2 H plus 5 divided by the square root of 25 + 2 H + 5. And we need to recall a note that A minus B multiplied by parenthesis A plus B is equal to A2 minus B2. We will achieve the following answer, which will be the limit as H approaches 0 of 25 + 2 H in parenthesis minus 25. Divided by H multiplied by the square root of 25 + 2 H + 5 in parenthesis. Fantastic. And when we do a little bit of simplifying, We should get, and I'm gonna skip a few steps, but when we simplify, we will get that the limit as H approaches 0 of 2 H divided by H multiplied by the square root of 2525 plus 2 H. So it's gonna be H multiplied by the square root of 25 + 2 H plus 5. Fantastic. So now, When we cancel out H, we will be left simply with The limit as H approaches 0 of. Drum roll, 2 divided by the square root of 25 + 2 H + 5. Fantastic. So now, our next step is we need to substitute in H equals 0, and when we do that, we will find that 2 divided by the square root of 25 + 5 is going to be simply equal to 15th. When we plug that into our calculator. So, therefore, without a doubt, our slope has to be 1/5. So let us note that M denotes slope, and our slope at X equals 9 is gonna be 1/5 or 1 divided by 5. Fantastic. So now our next step is we now need to solve for the tangent line. So as we should recall, in order to solve for the tangent line equation, we need to recall and use the point slope form, which, as we should recall, the point slope form states that Y minus Y1 is equal to the slope multiplied by X minus X1. So let us note that once again that M or slope is equal to 15th. Let us also note that X1 is equal to 9, and let us also note that Y1 is equal to 5. So when we plug in our known variables, we will get Y minus 5 is equal to 1/5 multiplied. By X minus 9. So when we rearrange this equation to isolate and solve for Y all by itself, skipping a few steps but using a little bit algebra, we will find that Y is equal to 1/5 X plus 16/5 or 16 divided by 5, and that is our final answer for the tangent line. Hooray, we did it. So Y is equal to 15 X plus 16/5, and that's it. We've officially solved for this problem. So going back up to the top to quickly recap what we've learned, our final two answers without a doubt have to be our slope is M equals 1/5, and our tangent line is Y equals 1/5 X plus 16 divided by 5. 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.

f(x) = √(x + 1), (8, 3)

Key Concepts

Derivative

Tangent Line

Point-Slope Form

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