Assume that functions f and g are differentiable with f(2) = 3...

Question

Assume that functions f and g are differentiable with f(2) = 3, f'(2) = −1, g(2) = −4, and g'(2) = 1. Find an equation of the line perpendicular to the line tangent to the graph of F(x) = (f(x) + 3) / (x − g(x)) at x = 2.

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says given that H and K are differentiable functions with H 3 equal to 2, H3 equal to 2, K 3 equal to 5, and K 3 equal to -2 determine the equation of the line perpendicular to the tangent line to the graph of capital H X, which is equal to the quantity of 2 multiplied by H X minus 4 in quantity, divided by the quantity of X plus k of X in quantity at X equal to 3. We're given 4 possible choices as our answers. For choice A, we have Y is equal to -2 X minus 6. For choice B, we have Y is equal to -2 X plus 6. For choice C, we have Y is equal to 6 X minus 2. And for choice D, we have Y is equal to 6 X plus 2. Now we need to determine the equation for the line perpendicular to the tangent line of the graph of capital H X at X equal to 3. The first thing we want to do is determine the value of capital H of X when X is equal to 3. So we need to calculate H of 3. Which is equal to the quantity of 2 multiplied by H of 3 minus 4 in quantity, divided by the quantity of 3 plus K of 3 in quantity. And if we look at the problem text, we see that we're given that H of 3 is equal to 2, and K of 3 is equal to 5, so we'll substitute in those values. So this is going to be equal to 2 multiplied by 2 minus 4, and that quantity is going to be divided by the quantity of 3 + 5. And when we evaluate this expression, this is going to be equal to 0. So that means our tangent line intersects our graph at the point of 3.0. Next, we need to um determine the slope of our tangent line when X is equal to 3, because we'll need the slope of our tangent line in order to determine the slope of the line that's going to be perpendicular to it. So, in order to determine the slope of the tangent line, we'll need to take a derivative of our graph, H H of X with respect to X. So we'll need to calculate H of X, and this is going to be equal to the derivative with respect to X. Of the quantity of 2 multiplied by H X minus 4 in quantity divided by the quantity of X plus K of X in quantity. And so here I'm taking a derivative of a fraction, so we want to use our quotient rule. So recall for the quotient rule that if we have a function Y that has the form, Y is equal to U X divided by V of X. Then its derivative, Y is equal to the quantity of V multiplied by U minus U multiplied by V in quantity divided by V2d. So, in our case, U is going to be equal to 2 multiplied by H X minus 4, and V is going to be equal to X plus K of X. And so that means that you is going to be equal to the derivative with respect to X of the quantity of 2 multiplied by H of X minus 4 in quantity, and taking this derivative, this is going to be equal to 2 multiplied by H of X. And for V, that's going to be equal to the derivative with respect to X of the quantity of X plus K of X. In quantity and taking this derivative, this is going to be equal to 1 plus K of X. So that means that our derivative capital H X. It's going to be equal to using the quotient rule, the quantity. of X plus K of X. In quantity, multiplied by the quantity of 2 multiplied by H of X in quantity, minus the quantity of 2 multiplied by H X minus 4. In quantity, multiplied by the quantity of 1 plus K of X in quantity, and all of that is divided by The quantity of X plus K of X. In quantity squared. So now we just need to find the slope of our tangent line when X is equal to 3. So we need to evaluate H of 3. So doing so, we'll have capital H of 3, and that's going to be equal to the quantity of 3 plus, and here we'll have K of 3, which is equal to 5. In quantity, multiplied by the quantity of 2, multiplied by H of 3, that's also equal to 2, so we'll multiply that by 2. In quantity, minus the quantity of 2 multiplied by H of 3, which is equal to 2. Minus 4 in quantity, multiplied by the quantity of 1. Plus K of 3, which is equal to -2. In quantity And all of that is divided by. The quantity of 3 + 5 in quantity squared. So, simplifying this expression, this is going to be equal to. 32 0, and that quantity is going to be divided by 64. And simplifying further, this is going to be equal to 1 divided by 2. So the slope of our tangent line is going to be equal to 1 divided by 2, but we're interested in the equation for the line that is perpendicular to this. So recall that if two lines are perpendicular, that their slopes need to be the negative reciprocals of each other. So the slope of the line that we're looking for, which we'll label as M. It's going to be equal to the negative reciprocal of 1 divided by 2, which is equal to minus 2. And the equation for this line must also pass through the point of 3.0. So here we have a slope and a point, and we can use that information to write the equation for our line using the point slope form. So we call for the point slope form that you'll have Y minus our particular Y value, which in this case is 0, and that's going to be equal to our slope, which is -2, multiplied by the quantity of X minus our particular X value, which in this case is 3, in quantity. And then distributing to and simplifying our expression here, we're going to get Y. is equal to -2 X plus 6. So, this is the equation that we were looking for. And we compare our equation for our line here to our answer choices, we see that the correct answer choice corresponds to answer choice B. So, I hope that this explanation has been useful, and I'll see you in the next video.

Assume that functions f and g are differentiable with f(2) = 3, f'(2) = −1, g(2) = −4, and g'(2) = 1. Find an equation of the line perpendicular to the line tangent to the graph of F(x) = (f(x) + 3) / (x − g(x)) at x = 2.

Key Concepts

Differentiation

Tangent and Perpendicular Lines

Function Composition and Quotients

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