Describe the concavity of the graph and find the points of inflection (if any)....?

Question

Describe the concavity of the graph and find the points of inflection (if any).

1) f(x) = x^3 - 3x + 2

2) f(x) = x^3 (1-x)

Do NOT use a graping calculator, show where the concavities are (and inflection points if there are any). Show your work. Thanks! :)

Answer 1

the slope of the tangent line tells you where a fcn is changing from increasing to decreasing or increasing to decreasing.

anywhere f '(x) is negative the fcn is decreasing
anywhere f '(x) is positive the fcn is increasing
the point at which the f '(x) switches sign is a local min or max

it's a max if f '(x) goes from + to -
it's a min if f '(x) goes from - to +
this also tells you the concavity of the fcn

points of inflection are kind of the middle between a min and max or a max and min. between the inflection points the graph may be con-up or con-dwn. again this depends on the sign of f '(x)

Answer 2

Hello,
1) f'(x) = 3x^2 -3 = 0 so x^2 = 1 and x = +/- 1.

Now f''(x) = 6x = 0 and x = 0

Since the graph increases as x < 0 and increases as x>= then the point (0,2) is a point of inflection and then (-1,4) is a relative minimum and (1,0) is a relative minimum.


2) f(x) = x^3 - x^4 f'(x) = 3x^2 -4x^3 set = 0 we have

x^2(3 - 4x) = 0 so x = 0 and x = 3/4

Now let's look at the graph on both sides of 0

so at -1 it has a value of -2 and at .1 (stay away from 3/4) it has a value of .0009 So it goes from negative to positive then at x = 0 it is a point of inflection and at x = 3/4 it is a relative max.

Hope This Helps!!