Calculus Help?

Question

Need help on this problem Thanks.

Use analytic methods to find the intervals where the function f(x)=-2X^3+6X^2-3 is:
a.increasing
b.decreasing
c.concave up
d.concave down
e.local extreme values
f.inflection points

Answer 1

f'(x)=-6x^2+12x
setting it to zero
-6x(x-2)=0
x=0 or 2
f''(x)=-12x+12
f'''(x)=-12
at x=0 there is a local min and at x=2 there is a local max
decreasing from - infinity to 0 and increasingfrom 0 to 2 and
decreasing from 2 to infinity
concave up at 0 and concave down at2
no inflexion points

Answer 2

f(x)=2x^3 +6x2 - 3
First derivative is 6x^2+12x = 6x(x+2)
Second derivative is 12x + 1

From the first derivative we see that f(x) has critical points at x=0 and x=-2. When x = -3, the first derivative = 6(-3)(-3+2)= 18. When x= -1, the first derivative = 6(-1)^2 +(12)(-1)= + 6. Thus as x passes through x= -2 from left to right its value changes from + to minus. This means it was positive at x = -3, 0 at x = -2 and positive at x = -1. Therefore x +-2 is a minimum of f(x) an so is decreasing for values of x < -2 and then is 0 at x=-2 and then starts increasing for x > -2.

Now the second critical point at x=0 should be tested the same way, For values of x less than 0 (say -1), the first derivative is positive and for values of x > 0 it is negative. Therefore f(x) has a maximum at x= 0 so f(x) increases over the interval -2
Summary:
- infinity < x< -2 [f(x) is decreasing]
-2 0 f(x) has a minimum at x = -2
f(x) has a maximum at x=0
f(x) is concave down at - infinity < x < 0
f(x) is concave up at - 0 < x < + infinity
The second derivative is 0 when x = -1/12 and is a point of a value of x where a point of inflection occurs, that is f(x ) changes concave down to concave up.