# Math Help - area limited by the curve

1. ## area limited by the curve

Hello

I need to find the plane area limited by the curve and the horizontal tangent line...

y = - 6x^2 + 5x^4

Thanks!

2. tangent line at what point on the curve?

"area limited by the curve" limited by what?

3. Originally Posted by Pommac
Hello

I need to find the tangent line and the area limited by the curve...

y = - 6x^2 + 5x^4

Thanks!
Hi

Some elements are lacking : tangent line at which point ? area limited by which abscissas ?

4. Originally Posted by skeeter
tangent line at what point on the curve?

"area limited by the curve" limited by what?
Sorry
... limited by the curve and the *horizontal tangent line

5. LOL

got news for you ... there are two distinct horizontal tangent lines to the given curve.

which one?

6. Originally Posted by skeeter
LOL

got news for you ... there are two distinct horizontal tangent lines to the given curve.

which one?
It actually says find the plane area limited by the curve and the horizontal tangent line.
Seriously I don't really know what that means.

7. $y = 5x^4 - 6x^2$

$y' = 20x^3 - 12x$

$y' = 4x(5x^2 - 3)$

one horizontal tangent line is $y = 0$ at $x = 0$

another horizontal tangent line, $y = -\frac{9}{5}$, touches the curve at two points ...

$\left(\sqrt{\frac{3}{5}} , -\frac{9}{5}\right) and \left(-\sqrt{\frac{3}{5}} , -\frac{9}{5}\right)$

here lies the dilemma ...

the horizontal tangent line $y = 0$ (the x-axis) bounds the curve above. there are two symmetrical regions formed by the curve below the x-axis.

$A = 2 \int_0^{\sqrt{\frac{6}{5}}} 6x^2 - 5x^4 \, dx$

second possible region ...

the horizontal line $y = -\frac{9}{5}$ bounds the curve below ...

$A = 2 \int_0^{\sqrt{\frac{3}{5}}} 5x^4 - 6x^2 + \sqrt{\frac{9}{5}} \, dx$

I recommend that you graph the original function on a calculator so you can visualize what I'm talking about.