In Exercises 47 and 48, find an equation for

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Question

In Exercises 47 and 48, find an equation for

(b) the horizontal tangent line to the curve at Q.

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Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says determine the equation of the horizontal tangent line to the following curve at point B. Below the problem we've given a graph that has the curve Y equal to 4 plus tangent of X minus 2 multiplied by CXx plotted on it. We also have a horizontal tangent line that intersects our curve at the point B. Now, we need to deter the equation of this horizontal tangent line, and the first step is to determine the derivative of our function Y. So, we'll need to calculate Y and that's going to be equal to the derivative with respect to X. Of quantity of 4 plus tangent of X. Minus 2 multiplied by sequant of X. In quantity So taking this derivative, this is going to be equal to. For the first term, we'll have the derivative of 4 with respect to X. So here I'm taking a derivative of a constant and recall that if I take a derivative of a constant, that that derivative is going to be equal to 0. So I'll have 0, plus for the second term, we'll have the derivative with respect to X of tangent of X, and so recall for that derivative, we have sequin squared of X. Then we'll have minus the derivative with respect to X of 2 multiplied by C X. The 2 is constant so we can move through the derivative, so we'll have 2 multiplied by the derivative with respect to X of C and the X, and we call for that derivative, that's going to be equal to sequent of X, multiplied by tangent of X. And so here we want to find the equation for a horizontal tangent line, and recall that a horizontal tangent line has a slope of 0. And so that means that the slope of that tangent line must be the same value as our slope of our curve Y, which is going to be equal to the derivative at point B. So that means we can set sequent squared of X minus 2 multiplied by sequent of X, multiplied by tangent of X. Equal to 0 and solve it for X, and this will give us the X coordinate of our point B. So, the first step to solve this equation is to add 2 multiplied by sequent of X, multiplied by tangent of X to both sides of the equation. Doing so, we'll get sequent squad of X. is equal to 2 multiplied by sequent of X, multiplied by tangent of X. Now, sequin of X is never equal to zero for any real value of X, so we can divide both sides of this equation by sequin of X. In doing so, we'll have sequent of X. Equal to 2 multiplied by tangent of X. So what we're going to do now is rewrite the sequin and tangent functions in terms of science and cosines. So we call that sequent of X is equal to 1 divided by cosine of X. And that's going to be equal to 2 multiplied by the quantity, and recall that tangent of X is equal to the sin of x divided by cosine of X. So now we'll multiply both sides of the equation by cosine of X, and doing so, we'll get 1 is equal to 2 multiplied by sin of X. Then we'll need to divide both sides by 2, so we'll get 1 divided by 2 is equal to sin of x. And now we just need to take the inverse sign of both sides of this equation. So doing so, we'll get X is equal to the inverse sign. Of the quantity of 1 divided by 2. Now this has a general solution of X is equal to pi divided by 6, plus 2 pi multiplied by k, where k here is an integer. However, if you look at our graph, we see that our point B falls between the X values of minus pi divided by 2 and pi divided by 2. So the only value of X that corresponds to that domain is when K is equal to 0. So that means that our X is going to be equal to pi divided by 6. So now that we have the X value for our coordinate B, we can use it to find its Y value. And so, that means that Y is going to be equal to 4 plus. Tangent of pi divided by 6. Minus 2 multiplied by sequant of pi divided by 6. So evaluating this expression we'll have Y is equal to 4 plus and tangent of pi divided by 6 is equal to 1 divided by the square root of 3. Then we'll have minus 2 multiplied by the quantity, and 2 of pi divided by 6 is equal to 2 divided by the square root of 3. So simplifying this expression further, we'll have Y is equal to 4, plus, and here we'll combine the last two terms with a common denominator of square root of 3. So we'll have 1 minus 4, which will give me minus 3, and that it's divided by the square root of 3. And simplifying this one more time, we'll have Y is equal to 4 minus 3 divided by square root of 3, which is square root of 3. And so this is the Y value for our point B, but this is also the equation for our horizontal tangent line. So our horizontal tangent line is horizontal and it just corresponds to a y value. So the equation for our horizontal tangent line is Y is equal to 4 minus the square root of 3. And so that is going to be the answer for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

In Exercises 47 and 48, find an equation for

(b) the horizontal tangent line to the curve at Q.

Key Concepts

Horizontal Tangent Line

Derivative of Trigonometric Functions

Critical Points

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