SOLUTION: Consider this function f(x) = x^3+ax^2+bx+c where a>=0, b>0
Over what intervals is f concave up? Concave down?
Show that f must have one inflection point.
Given that (0,-2) is t
Question 1027710: Consider this function f(x) = x^3+ax^2+bx+c where a>=0, b>0
Over what intervals is f concave up? Concave down?
Show that f must have one inflection point.
Given that (0,-2) is the inflection point of f, solve for a and c and then show that f has no critical points. Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website!
==> f'(x) = ==> f"(x) = 6x + 2a
Setting f" to zero, we get .
To the right of -a/3, f" >0, while to its left f" < 0. (a > 0!)
==> To the right of -a/3, f is concave up, while to its left f is concave down.
Hence is an inflection point of f.
If (0,-2) is an inflection point then , and so a = 0.
==> ==> ==> c = -2
==> .
==> f'(x) = .
Since f' > 0 for all x because b > 0, f will not have any critical points, and the problem is solved.
