Integral Calculus

Question

Find a function y = ƒ(x) having atleast three inflection points and sketch it's graph. [ Note: To find the formula of ƒ(x) start with a suitably chosen ƒⁿ(x) and integrate ƒⁿ(x) twice. The second part of the question is a curve sketching problem. Give all details. ]

Answer 1

Look Hamid F.,

The question specifies exactly what you should go about doing!. Inflection points arise from 0-crossings of the second derivative of a function, right?

So you want a second derivative function with 3 zeros. And by the way, you meant "its graph", not "it's graph".

To get 3 zeros, just multiply three different terms, such as: (x+2), (x-2), (x-4). You have a 3rd degree polynomial. Integrate it twice vs dx, throw in two arbitrary constants if you want, and voila`!

Don't forget the excruciating details.

Make sure you show considerably more than just -2 to +4 region of x, otherwise you can't possibly graphically see the inflection points, and you might miss some local maxima and minima, too.

-- Cheers, --Mr. d.