How?Physics1 said:Homework Statement
y = 9sin(x)cos(x)
Find all points where tangent line is horizontal.
The Attempt at a Solution
I get y' = 9cos^2x - 9sin^2x
I plug in zero for the slope and get 9 but I'm stumped after that.
You do exactly what you say to do; that is, find the points where the derivative of the graph is zero. It boils down to solving cos^2x-sin^2x=0. Can you do this?How can I get all the horizontal tangent line points?
The equation for finding horizontal tangent points is to set the derivative of the function equal to 0. In this case, the function is y = 9sin(x)cos(x), so you would set its derivative, which is y' = 9cos^2(x) - 9sin^2(x), equal to 0.
To solve for x, you will need to use algebraic techniques. In this specific example, you can factor out a 9 from both terms on the right side of the equation, giving you 9(cos^2(x) - sin^2(x)) = 0. Then, you can use the trigonometric identity cos^2(x) - sin^2(x) = cos(2x) to rewrite the equation as 9cos(2x) = 0. Finally, divide both sides by 9 and take the inverse cosine of both sides to solve for x.
There are an infinite number of horizontal tangent points for this function. This is because the cosine and sine functions are periodic and will repeat themselves infinitely.
Finding horizontal tangent points can help you determine the maximum and minimum values of a function, as well as the points where the function changes from increasing to decreasing or vice versa.
Yes, you can use a graphing calculator to find horizontal tangent points. Most graphing calculators have a feature that allows you to find the zeros of a function, which in this case would correspond to the horizontal tangent points.