Question

Find the x-coordinate of the point(s) of inflection on the graph of y = x^3 – 6x^2 + 9x + 3.
a. x = 0
b. x = 1
c. x = 2
d. x = 3

Answer 1

the critical pts are 1 and 3

Answer 2

@ganeshie8 @Directrix @mathmate @AuroraB @iamawesome1 @Abhisar @wio @iGreen @nincompoop @Preetha @uri @TheSmartOne

Answer 3

@Michele_Laino @AlexandervonHumboldt2 @Abhisar

Answer 4

Do I find the second derivative?

Answer 5

f'(x)= 3x^2-12x+9
f''(x)= 6x-12

Answer 6

If I set 6x=12 then x=2

Answer 7

\[find \frac{ d^2y }{ dx^2 }~at~x=1,x=3\]
where \[\frac{ d^2y }{ dx^2 }=0\]
that point is point of inflexion

Answer 8

from wht u told me to do i got (1,-6)(3,6)

Answer 9

@surjithayer

Answer 10

so now what do i do?

Answer 11

at x=2, f"(x)=0
so point of inflexion is x=2
1. at point of inflexion concavity or convexity changes

Answer 12

so the answers 2 then?

Answer 13

@paki @ganeshie8 @Directrix @mathmate @AuroraB @iamawesome1 @wio @iGreen @nincompoop @Preetha @uri @TheSmartOne @surjithayer @Abhisar @Alexistheamazing @AlexandervonHumboldt2 @Quan99 @wio @Loser66 @Compassionate @uri