Question

Determine the intervals on which f(x)=-x^4–2x^3+12x^2
are concave up or concave down. Identify any inflection points.

Answer 1

find the 1st and 2nd derivatives


set the 1st derivative to zero and solve for x, this will give any stationary points

test the points in the 2nd derivative to determine their nature

set the 2nd derivative equal to zero and solve.

check for a change in concavity either side of these solutions.

these will be the points of inflexion.

concave up occurs between the maximum values

concave down occurs between the minimum values...

Answer 2

lol sorry i know that.. i was trying to type my answers on my phone but it didn't work lol

i got C.up @ (-2,1) c.down @ (-inf, -2) , (1, +inf) and inflection poits @ x=-2 and x=1

i don't have the answers on the book so i need to know if I'm correct

Answer 3

@campbell_st

Answer 4

well here is a graph that may help

Answer 5

yeeep ok .. i forgot i could look in graphs lol .. thanks :D

Answer 6

well I these are my derivatives

and the stationary points aren't nice as the 1st derivative is

\[y' = -4x^3 -6x^2 + 24x \]

or

\[y' = -2x(2x^2 + 3x -12)\]

and you'd need the general quadratic formula to solve the quadratic... which is messy

Answer 7

and then the 2nd derivative is

\[y" = -12x^2 -12x + 24\]

which is

\[y" -12(x^2 + x -2)\]

and a lot easier to find the points of inflexion..

Answer 8

yea i did the first derivative and saw that with the quadratic form. thing weren't coming out nice so i went straight to the 2nd deriv. and well, by looking at the graph you gave me, i think i got the answers right

Answer 9

can i ask you another quick question? @campbell_st

Answer 10

when i sketching a curve..
and it says f'<0 and f''>0, for -1<x<2
that means that
the first derivative is negative and
the second derivative is positive
before -1 and after +2
correct?

Answer 11

nevermind i got the answer