When is the ORIGINAL FUNCTION concave up, if f'(x)=(x-2)^2(x-3)^2?

Answer 1

If that's f'(x), then the f(x) term will have a x^5 term in it, which means that the original function will always be concave up somewhere in the curve.

Answer 2

To find the inflection points of concavity of f(x), first you have to take the first derivative f'(x) that you've stated and differentiate again to find the second derivative f''(x)... to save you time it's [2(x-2)(x-3)(2X-5)].. Then you have to find the points in f''(x) which yield either 0 or are undefined (i.e., some number divided by zero). Since this function is not rational (or given as a fraction divided by another fraction), it will never be undefined, but it will yield zero at the points x=2, x=3, and x=(5/2). From here, you plug in four values to f''(x) that are between and outside these critical points - one that is <2, one that is >2 but <(5/2), one that is >(5/2) but <3, and one that is >3. The results will be the same no matter what "in-between values" you choose for each interval. Where the intervals are positive, the original function f will be concave up... in this case it will be be on the intervals (2, (5/2)) and (3, infinity). Hope this helps.

Answer 3

f''(x) = 2(x-2)(x-3)^2 + 2(x-2)^2(x-3) = 2(x-2)(x-3)(2x-5) = 0

Solve for x,
x =2, 2.5, 3

On (-infinity, 2), f"(x) < 0. f(x) is concave down.
On (2, 2,5), f"(x) > 0. f(x) is concave up.
On (2.5, 3), f"(x) < 0. f(x) is concave down.
On (3, infinity), f"(x) > 0. f(x) is concave up.