Hello
I need to find the plane area limited by the curve and the horizontal tangent line...
y = - 6x^2 + 5x^4
Thanks!
$\displaystyle y = 5x^4 - 6x^2$
$\displaystyle y' = 20x^3 - 12x$
$\displaystyle y' = 4x(5x^2 - 3)$
one horizontal tangent line is $\displaystyle y = 0$ at $\displaystyle x = 0$
another horizontal tangent line, $\displaystyle y = -\frac{9}{5}$, touches the curve at two points ...
$\displaystyle \left(\sqrt{\frac{3}{5}} , -\frac{9}{5}\right) and \left(-\sqrt{\frac{3}{5}} , -\frac{9}{5}\right)$
here lies the dilemma ...
the horizontal tangent line $\displaystyle y = 0$ (the x-axis) bounds the curve above. there are two symmetrical regions formed by the curve below the x-axis.
$\displaystyle A = 2 \int_0^{\sqrt{\frac{6}{5}}} 6x^2 - 5x^4 \, dx$
second possible region ...
the horizontal line $\displaystyle y = -\frac{9}{5}$ bounds the curve below ...
$\displaystyle A = 2 \int_0^{\sqrt{\frac{3}{5}}} 5x^4 - 6x^2 + \sqrt{\frac{9}{5}} \, dx$
I recommend that you graph the original function on a calculator so you can visualize what I'm talking about.