a. Show that dy/dx= (3y-2x)/(8y-3x)
2x+8y dy/dx=3y+3x dy/dx
(ex+8y)dy/dx=3y-2x
dy/dx=(3y-2x)/(8y-3x)
b. Show that there is a point P w/ x-coord. 3 at which the line tan to the curve at P is horizontal. Find y-coord of P.
(3y-2x)/(8y-3x)=0
3y=2x^3
3y=6
y=2
(3,2)
c. Find value of d^2y/dx^2 at point P in B. Dos curve have local max, min, or neither? justify
[ (3dy/dx-2)(8y-3x)-(8dy/dx-3)(3y-2x)]/(8y-3x)^2
=-14/49
=-2/7
because y second deriv is neg, the curve is concave down. Therefore there is a max at (3,2)