Tabris
Member
- Joined
- Mar 16, 2004
- Messages
- 806
- Gender
- Male
- HSC
- 2004
Consider the curve y = x^4 - 4x^3
find coordinate of stat points
determine anture of these stat points
sketch the curve using the above information
for what values of x is the curve concave up
x^4 - 4x^3
y' = 4x^3 - 12x^2
y"= 12x^2 - 24x
for stat points
4x^3 - 12x^2 = 0
4x^2(x-3) = 0
x = 3 and 0 meaning (3,27) and (0,0)
at x = 3
y">0
at x = 0
y" = 0 (pt of inflection)
when the point of inflection is ar the origin i tested it using -0.5, 0, and o.5
when subbed into y', -0.5 and 0.5 both got a negative gradiant meaning horizontal point of inflection right?
12x^2 - 24x > 0
12x^2 > 24x
x >2 (divided both sides by 12x)
so therefore the curve is concave up when x > 2
but how could that be if the state point is at (3,-27)?
anyone know where i made a mistake? ta, thanks for your time
find coordinate of stat points
determine anture of these stat points
sketch the curve using the above information
for what values of x is the curve concave up
x^4 - 4x^3
y' = 4x^3 - 12x^2
y"= 12x^2 - 24x
for stat points
4x^3 - 12x^2 = 0
4x^2(x-3) = 0
x = 3 and 0 meaning (3,27) and (0,0)
at x = 3
y">0
at x = 0
y" = 0 (pt of inflection)
when the point of inflection is ar the origin i tested it using -0.5, 0, and o.5
when subbed into y', -0.5 and 0.5 both got a negative gradiant meaning horizontal point of inflection right?
12x^2 - 24x > 0
12x^2 > 24x
x >2 (divided both sides by 12x)
so therefore the curve is concave up when x > 2
but how could that be if the state point is at (3,-27)?
anyone know where i made a mistake? ta, thanks for your time