Requesting Help with Calculas Solution.

Basic

Hello again guys, I am in need of your help again.

Find the intervals of concavity and the inflection points:
a) f(x) = 2x^3 - 3x^2 - 12x

Solution: f''(x) = 6(2x - 1) => f''(x) > 0 on (1/2, infinite) and f''(x) < 0 on (-infinite, 1/2). So f is concave upward on (1/2, infinite) and concave downward on (-infinite, 1/2). There is a change in concavity at x = 1/2; and we have inflection point at (1/2, -13/2).

My problem: How does one get the above (in bold)? It escapes me and I would really appreciate it if someone helped me out here.

Clipse

f''(x) is the second derivative of f(x), which is the derivative of the derivative of f. Take the derivative of f to get f'(x) = 6x^2 - 6x - 12, and take the derivative of that to get f"(x) = 12x - 6 = 6(2x-1). (Both are just power rule derivatives - I'm assuming you're familiar with that.)

Basic

Clipse wrote: »

f''(x) is the second derivative of f(x), which is the derivative of the derivative of f. Take the derivative of f to get f'(x) = 6x^2 - 6x - 12, and take the derivative of that to get f"(x) = 12x - 6 = 6(2x-1). (Both are just power rule derivatives - I'm assuming you're familiar with that.)

This makes it clear to me, thanks a lot, really.

FrostyAlphaWolf

If you're just looking for the bolded section, it seems like taking the derivative will get you there in no time.

If you're curious to the whole process, it's not too much more complicated. To find the concavity, you'll need to find the inflection points (aka, critical points). To do so, take the first derivativeof the original equation. Then, when you have f'(x), set the equation to zero. The values you find for x from that process are your test points.

Take the derivative of f'(x), giving you f''(x); once you have this new equation, plug in each of your test points, one at a time, into the equation. Compare the results for each case to see if the test point is a valid inflection point, or if it is invalid due to being out of the given domain. Once you have gone through all your test points, you're halfway done.

Write out the newly found inflection points (helps to make a small table). You will then need to test points greater and less than each of the points in f''(x). If the solution for a particular value is positive, the concavity is upward (facing upwards, "collecting water"); however if it is negative, the concavity is downward (facing down, "pouring water").

Big Dookie

Basically, to find Minima, Maxima, and Inflection points, you need to use the second derivative test. First, you find your first derivative - which in your case is f'(x) = 6x^2 - 6x - 12, and then use this to find your second derivative, f''(x) = 12x - 6, or f''(x) = 6(2x-1). Here's the rule for second derivatives:

1) If f''(x) > 0 then (x , f(x)) is a minimum.
2) If f''(x) < 0 then (x , f(x)) is a maximum.
3) If f''(x) switches signs at x then (x , f(x)) is an inflection point.

So basically, all you do is find what values of x will give you an f''(x) that is greater than zero, find what values of x will give you an f''(x) less than zero, and where f''(x) equals zero (or where it switches signs, basically). If it's a minimum, that means the function between those x values concaves downward, and if it's a maximum, it means the function between those x values concaves upward. Where it changes in concavity is the inflection point.