(pretty basic) Calculus help?

the first thing

(x^3/3) - (x^2/2)

((1/3)*x^3)-((1/2)*x^2)

lower the powers by 1, then times the old power by the things in front (1/3 and 1/2)

((3/3)*x^2) - ((2/2)*x)

x^2 - x

Second thing

so you differentiate it, then, do this:

0=the derivative

solve it for x

sub x into the original equation

solve for y

now you have the x and y co-ords of the turning point

y= (x^3/3) - (x^2/2)

What's d/dx(x^n)? It's n*x^(n - 1)

Thus dy/dx = (1/3)*3*x^2 - (1/2)*2*x

= x^2 - x = x*(x - 1) <<<

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y=x^3 + 9x^2 + 15x + 1

Turning points are where dy/dx = 0

dy/dx = 3*x^2 + 18*x + 15

= 3*(x*2 + 6*x + 5)

=3*(x + 1)*(x + 5)

= 0 if x = -1 or x = -5

x = -1, y = -1 + 9 - 15 + 1 = -6

x = -5, y = -125 + 225 - 75 + 1 = 24

Thus the turning points are (-1.-6) and (-5,24) <<<

(Check my arithmetic!)