Question

This is a third degree polynomial

y=(x-1)(x-2)(x-5)

so it has roots (y=0) at (x=1,2 & 5) and has two points where the slope is zero , one between x=1 and x=2 and the other between x=2 and x=5.

Differentiation provides a means to locate those points.

using this rule of differentiation

since y=f(x)g(x)h(x)

dy/dx=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)

and setting dy/dx=0 (slope is zero)

you can solve for the 2 x's

now to specifics

since

y=(x-1)(x-2)(x-5)

dy/dx=(x-2)(x-5)+(x-1)(x-5)+(x-1)(x-2)=0

or dy/dx=3x^2-16x+17=0

solving the derivative for zeros

using the quadratic eqn [-b+/-sqrt(b^2-4ac)]/(2a)

or x=[8+/-sqrt(13)]/3

so there is one inflection point at x=[8-sqrt(13)]/3 and another at x=[x+sqrt(13)]/3

so go back to the original eqn

y=(x-1)(x-2)(x-5) and plug in these two x's and solve for y's note--these become your desired coordinate (x,y) pairs.

Now are these inflection points local maxima or minima?---an easy way to determine this is to determine where the location of ends of the cubic polynomial. So if at x>5 is y>0? It is so the inflection point (dy/dx=0) between x=2 and x=5 is a local minima. And xince a cubic equation tails off in the opposite direction then the other inflection point is a local maxima.

Rap