Find the coordinates of all points where the curve $y=rac{1}{x^{2}+x+1}$ has a horizontal tangent line....

Find an equation of the tangent line to the curve at the given point.

$ y = \sqrt{x} $,

$ (1, 1) $

Suppose you want to find an equation of the tangent line to the curve. Why? Which is equal to two? X plus one over X plus two. At the .1, 1 to do this, we first find the slope of the tangent line. That is the derivative of the function Evaluated at the given .1, 1. Now by quotient rule we have white prime, that's equal to X plus two times the derivative of the numerator two, x plus one-. We have to express one times the derivative of the denominator which is expressed to this all over the square of the denominator. And then from here we have X plus two Times derivative of to Express one. That's just too minus two, X plus one times the derivative of X plus two which is one. And then this all over the square of X plus two. And simplifying this, we have two, X plus four minus two, X -1. This all over X plus two squared or this is just three over the square of experts to And so the slope of the tangent line at the .11 is given by that's dy over dx evaluated at 1 1. This is just three over one plus two squared. That's just 3/9 or 1/3. So this is the slope of the tangent line at 11. And then the next step would be to use the point slope formula of a line to find the equation of the tangent line. Now the point slope formula of a line states that the the equation of the line is just why minus? Why is that one? This is equal to the slope mm Of that line times X -X. Sub one. Since you already have our point Except one White 1 Which is just 11. And we have our slope, we found out To be won over three. Then the equation of the tangent line is just Why -1 That's equal to 1/3 times x -1. And simplifying this, we have why that's equal to 1/3, X minus 1/3 Plus one. Or that why is equal to 1/3 Plus 2/3. And so this is the equation of the tangent line at the point 11

Hey, it's Claire is the one you married here, So we're first gonna differentiate her function. We got six x square plus six X minus 12. We're gonna study equal to zero to find where the points where the tension line is horizontal. You get six x plus two times X minus one. So we see that X bias are negative, too. And one, we plug this in to our original equation to negative two comma 21 and one common negative six.

He It's clear. So when you right here. So our first gonna differentiate using the quotient role. So we have tea. Why? Over t x is equal to one plus each The x the derivative of one plus X minus one plus eat x The derivative times one plus x This is all over one plus Eat the X square. This is equal to one plus eat the X times one minus Eat the x one plus X All over one plus Eat the X square. This becomes equal to one minus x Eat the X over one plus each the X square and to find the slope of the 10 gents, we're gonna plug in zero when we get 1/4 and the equation becomes why minus one have equals 1/4 times X minus zero and this becomes equal to x over four plus 1/2

Our goal here is to find the equation of the tangent line to the curve. Y equals two to the X at the points you're a one. The slope of the tangent line will be the derivative at that point, So let's find the derivative. The derivative of two to the X is two to the X Times natural log tube, so the derivative evaluated at zero will be two to the zero power types. Natural log to and to to the zero power is one. So we have one times natural log to which is natural log to So that is the slope of the tangent line. Now let's use the point slope form. Why minus y one equals m times X minus X one to find the equation of the line. So we substitute one in for why one? Since that's the Y coordinate of our point. We substitute natural log to infer the slope, and we substitute zero in for X one. Since that's the X coordinate of our point, and now we simplify. So we have y minus one equals natural log to Times X, and then we can add one to both sides. And the equation of the Tanja line is why equals the natural log of two times X plus one