area of triangle formed by pts on curve where tangents.....

[DianaS]

Question: Find the area of the triangle formed by the 3 points on the curve y = ( x[sup:13ojp1ko]2[/sup:13ojp1ko] - 1 / x[sup:13ojp1ko]2[/sup:13ojp1ko] + 1)[sup:13ojp1ko]2[/sup:13ojp1ko], where the tangent lines are horizontal.

I don't know where to start this problem. I think I should first take the deivative of the function, probably using Chain Rule combined with the Quotient Rule, but I'm not sure. And then, when I get the points, how should I put them into the triangle are formula ( bh / 2) ?

Thanks in advance,
Diana

 

[stapel]

Question: Find the area of the triangle formed by the 3 points on the curve y = ( x[sup:2mz5i1yu]2[/sup:2mz5i1yu] - 1 / x[sup:2mz5i1yu]2[/sup:2mz5i1yu] + 1)[sup:2mz5i1yu]2[/sup:2mz5i1yu], where the tangent lines are horizontal.

What you have posted means the following:

. . . . .y = [x[sup:2mz5i1yu]2[/sup:2mz5i1yu] - (1/x[sup:2mz5i1yu]2[/sup:2mz5i1yu]) + 1][sup:2mz5i1yu]2[/sup:2mz5i1yu]

Was this what you meant?

I don't know where to start this problem.

You are asked the find the points on the curve where the tangents are horizontal. Use what you've learned about the relationship between the derivative of a function and the slope of its tangent lines, in order to find those points.

What you'll need to do after that will depend upon the points you find, and their orientation. And that will depend upon what you mean the function to be. So I'm afraid I can't get any further, with just the information provided thus far. Sorry.

Eliz.

 

[galactus]

You can use the quotient rule or even the product.

\(\displaystyle \left(\frac{x^{2}-1}{x^{2}+1}\right)^{2}=\frac{(x^{2}-1)^{2}}{(x^{2}+1)^{2}}\)

Quotient rule:

\(\displaystyle \frac{(x^{2}+1)^{2}(2)(x^{2}-1)(2x)-(x^{2}-1)^{2}(2)(x^{2}+1)(2x)}{(x^{2}+1)^{4}}\)

\(\displaystyle \frac{4x(x^{2}-1)(x^{2}+1)^{2}-4x(x^{2}-1)^{2}(x^{2}+1)}{(x^{2}+1)^{4}}\)

Can you simplify and continue?. After you have your derivative, set to 0 and solve. You should get 3 solutions. These are the vertices of your triangle which you can use to find the area.

Like I said, you can also use the product rule by rewriting as \(\displaystyle (x^{2}-1)^{2}(x^{2}+1)^{-2}\)

That will be even easier. Try both and see if you get the same thing. You should.

 

[Dr. Flim-Flam]

f(x) = [(x^2-1)/(x^2+1)]^2, f ' (x) = [8x(x^2-1)/(x^2+1)^3]. Setting the slope = to zero, we have

x = 0, f(0) = 1, x = 1,f(1) = 0, x = -1, f(-1) = 0. Hence we have the points (0,1),(1,0), and (-1,0)

Area of triangle = (1/2)*bh = (1/2)*2*1 = 1 square unit.