using second derivative

can someone explain for question b, in the solution why did the textbook say x/< 0, shouldn't it be x>/0 from solving 6x(2x-3)>/0

Screenshot 2025-03-03 003916.png

(edited 11 months ago)

Reply 1

relativq

if you see the graph, it's above y=0 at x=0 and left, and x=1.5 and right

Reply 2

sdfj

Original post

by Noname60

can someone explain for question b, in the solution why did the textbook say x/< 0, shouldn't it be x>/0 from solving 6x(2x-3)>/0
Screenshot 2025-03-03 003916.png

Well, first of all, you know that if f''(x) is positive, f(x) is convex, and if f''(x) is negative, f(x) is concave.

It's probably best to approach the question by finding the x values where f''(x) is under the x-axis, and the x values that the f''(x) is above the x-axis.

So 6x(2x-3)=0, x(12x-18)=0, x=0, x=3/2 - as I'm sure you know, those are the roots of f''(x). And it's sort of obvious given that the coefficient of the x² term is positive that the region under the x-axis for f''(x) is found where: 0<x<3/2

So f''(x) is negative (f(x) is concave) between 0/<x/<3/2 (the answer to part i)

AND f''(x) is positive (f(x) is convex) when x/<0, x>\ 3/2 (the answer to part ii)

Hope that makes sense!