Hello, NAPA55!
Here's the last one . . .
5. Find the area of the triangle formed by the three points
on the curve: .$\displaystyle y\:=\:\left(\frac{x^21}{x^2+1}\right)^2$ where the tangent lines are horizontal. First, find the derivative (Chain Rule, Quotient Rule, etc.)
. . $\displaystyle y' \;=\;2\left(\frac{x^21}{x^2+1}\right)\cdot\frac{(x^2+1)\cdot2x  (x^21)\cdot2x}{(x^2+1)^2} \;=\;\frac{8x(x^21)}{(x^2+1)^3}$
Tangents are horizontal where $\displaystyle y' = 0$
. . So we have: .$\displaystyle 8x(x1)(x+1) \:=\:0\quad\Rightarrow\quad x \;=\;0,\:\pm1$
When $\displaystyle x = 0\!:\;\;y \:=\:\left(\frac{01}{0+1}\right)^2 \:=\:1\quad\Rightarrow\quad (0,\,1)$
When $\displaystyle x = 1\!:\;\;y \:=\:\left(\frac{11}{1+1}\right)^2 \:=\:0\quad\Rightarrow\quad (1,\,0)$
When $\displaystyle x = \text{}1\!:\;\;y \:=\:\left(\frac{11}{1+1}\right)^2\:=\:0\quad\Rightarrow\quad (\text{}1,\,0)$
And surely you can find the area of this triangle . . .
Code:

*(0,1)
/  \
/  \
/  \
 *    +    * 
(1,0)  (1,0)