Find an equation of the line tangent to the graph of f at (a, ...

Question

Find an equation of the line tangent to the graph of f at (a, f(a)) for the given value of a.

f(x) = √x+2; a=7

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the equation of the tangent line to the curve of G of X equals the square root of X + 5 at X equals 4. A says it's Y equals -16 X minus 7/3, B-16 X plus 7/3, C 16 X minus 7/3, and D, 16 X plus 7/3. Now how can we figure out the equation of the tangent line? What do we know? Well, recall that we can write the equation in the point slope form that says Y minus Y1 equals m multiplied by X minus X1, where M is the slope of Y tangent line and X1 Y1 is a point on our line. Now in this case, before we can solve that means we'll need the slope and we'll need the point. Now we can find the slope, OK, because our slope M. is going to be equal to the derivative of our function or of our curved geoff X at the point where X equals 4. So first we'll need to find the derivative of geo X, plug 4 into the derivative to get the slope. And then we can get the point on our tangent line by using the XY point on the curve G of X, where our tangent meets our curve, OK? So first, let's focus on our slope. Let's start by finding the derivative of G of X. We can use the limit definition of the derivative. No, Gison, the GFX. Being equal to the square root of X + 5, OK. Then we know that the derivative of G of X by the limit definition is going to be equal to the limit as H approaches 0 of G X plus H minus G of X all divided by H. Now what are those going to be? Well, to find G X plus H, we're going to substitute X + H wherever we see X in G of X, OK? So now that's gonna be the limit as H approaches 0 of the square root of X + H+ 5 minus G of X, the square root of X + 5, all divided by H. Now if we notice here. To simplify our expression, if we substitute H equals 0 directly, then our denominator, then our function will be undefined because our denominator will be 0. So let's try to simplify this expression by rationalizing our numerator, OK? So that means we can multiply this by the square root of X + H+ 5 plus the square root of X + 5. And divided by the same term, square root of X plus H + 5 plus the square root of X + 5. No, when we do that, we'll be able to rationalize our enumerator, OK? So that we're multiplying the difference of two squares in the numerator, which will leave us with X plus H + 5 minus X minus 5. You know, expression, OK, and remember that for this we're finding the limit as it approaches 0. Let's not forget that. And in our denominator, it's going to be H multiplied by the square root of X + H+ 5, OK, plus I got properly here, the square root of X + 5. Knowing our numerator, X minus X is 05 minus 5 is 0, and we can simplify this to be the limit as H approaches 0. OK, of each divided by Or or or or original denominator from the previous part of our problem. Let me just copy and paste that for uh in the interest of time. No, H cancels H, and this is going to be equal to 1 divided by the square root of X plus H + 5 plus the square root of X + 5. And remember, don't forget that we're finding the limit. Myself have to remind myself of that. The limit as age approaches 0. Now we can substitute H as 0 in our expression. So that's going to be equal to 1 divided by the square root of X + 0 + 5 plus the square root of X + 5, OK? And X + 0 + 5 will be X + 5. And if we're adding the square root of X + 5 twice, this is going to be equal to 1 divided by 2 multiplied by the square root of X + 5. This is the derivative of G of X. Now that we have the derivative, we can evaluate it at X equals 4 to find the slope M. Because no at X equals 4. Then the derivative. Is going to be equal to 1 divided by 2 multiplied by the square root of 4 + 5. That's 1 divided by 2 multiplied by the square of 9. The square root of 9 is 3, and 2 multiplied by 3 equals 6. So at X equals 4, the slope M equals 16. Now that we have the slope, next, let's try to find a point, OK? We also know that at X equals 4, OK. Then G of X will be equal to the square root of 4 + 5, which equals the square root of 9, which is 3, OK? And rather I should rewrite this as G of 4, OK. So now, this also tells us that 43 is a point, OK. On the graph. And thus we can call 4 X1 and 3Y1. Now think about it. We have our slope M. We have X1 and Y1. No, we can substitute this into the point slope form of our equation. So no, this means that we can rewrite our equation as Y minus Y1 3 equal to M16 multiplied by X minus X1, which is 4. No, if we were to expand here, OK, we're gonna have Y minus 3 being equal to 1/6 of X minus 4/6 or 2/3, OK. And no, we're gonna get Y then to be equal to 16 of X minus 2/3 plus 3. Minus 2/3 + 3 equals 7/3, so this means Y is going to be equal to 1/6 of X. Plus 7/3 this is going to be the equation of our tangent line. Now, when we look back on our answer choices, OK, that tells us that D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Find an equation of the line tangent to the graph of f at (a, f(a)) for the given value of a.f(x) = √x+2; a=7

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Find an equation of the line tangent to the graph of f at (a, f(a)) for the given value of a.
f(x) = √x+2; a=7