Help !!! (1 Viewer)

[classact]

Please help, these questions are killing me, Ive done like 12 but these ones, are confusing me… Need by Friday Morning. Thanks Guys

Consider the curve given by y=x^3 -12x+4

a) find the co-ordinates of any stationary points and determine their nature
b) find the co-ordinates of any points of inflexion
c) for what value(s) of x in the domain -3≤x≤4 does y have its maximum value ?

a) What is the condition (in terms of dy/dx) for a function to be decreasing
b) Find the values of x for which the function y=4 + 36x-3x^2 – 2x^3 is decreasing

­­­Find
i) ∫(1-x^2)dx
ii) ∫(x + 1/x^2) dx

Show that ∫4 √2x + 1dx = 8 and two thirds.
0

And Finally …
The region, enclosed by the parabola y^2=4ax and the line x=a is rotated about the x axis. Find the volume of the solid formed.

Thank sooo much

 

[alexvincent]

You should probably post this in the maths subforum.

1a) y' = 3x^2 -12
stat pts when y' = 0
ie 3x^2 = 12
x^2 = 4
x = 2 or x = -2
at x = 2, y = -12 at x = -2, y = 20
so the two stat pts are A(2, -12) and B(-2, 20)
y" = 6x
at x = 2, y" = 12 > 0 hence A is a min
at x = -2 y" = -12 < 0 hence B is a max

b) pts of inflexion when y" = 0
6x = 0 ----> x = 0
at x = 0, y = 4
pt of inflexion (0, 4)

c) from part (a), there is only one maximum for the whole curve
hence in -3≤x≤4 there is also only one maximum ie the same max at x = -2

2a) dy/dx is the gradient function
so if the function is decreasing the gradient is negative
ie dy/dx < 0
b) dy/dx = 36 - 6x - 6x^2
for y to be decreasing
0 > 36 - 6x - 6x^2
solve for x

You really should be able to do the integration. Just look back in the text for the integration rules. Maybe revising your index laws will help too. Remember that breaking up fractions can be helpful.