Question

What are the points of inflection, if any, of `f(x)=x^4-10x^3+24x^2+3x+5`?

Answer 1

POI : (1, 23) and (4, 17). The graph is not to scale, Ad hoc scales enable location of the tangent-crossing-curve POI

Explanation:

#f''=12(x^2-5x+4)=0, at x = 1 and 4.

As f'''= 24x-60 ne 0, at these pints, the points of inflexion are at

(1, 23) and (4, 17)

graph{(x^4-10x^3+24x^2+3x+5-y)((x-1)^2+(y-23)^2-.17)((x-4)^2+(y-17)^2-.05)=0 [-10, 10, -60, 60]}