Differentiation?

Question

Can anyone explain why the second derivative at max <0 and at min it is > 0

Answer 1

A 2nd derivative < 0 tells you that the slope of the 1st derivative is decreasing as x increases, indicating concavity downwards, so when the 1st derivative passes through 0 it must be at a maximum. If the 1st derivative is negative and the 2nd positive you are approaching a minimum because the slope of the 1st derivative is increasing when it passes through 0.

Answer 2

Sure. The second derivatives tell you points of inflection (if I remember correctly). A minimum simple tells you where the graph is (as read from left to right) decreasing then increasing and a max tells you when it goes from increasing to decreasing. It doesn't tell you the actual lowest or highest point. It's slightly confusing at first, but a min can most definitely be more than a max.

Answer 3

first of all climb to the max. move a little to the left. what about the derivative at this point? it's positive (+) here. and what about the derivative at a point on the right of max? it is negative (-). so second derivative = { (-) - (+) } / dx = negative. similarly, for min.

Answer 4

Approaching a MAXIMUM turning point of a curve, the gradient is positive ie f `(x) = dy/dx is +ve and shape of curve is of form / sloping upwards

At the turning point, the gradient = 0 and shape of curve is of form -- horizontal

Immediately past the turning point the gradient is -ve
ie d/dx(dy/dx) = d²y/dx is -ve and the shape of the curve is \ sloping downwards

Similarly approaching a MINIMUM turning point, the gradient is - ve so shape \ sloping downwards

At the turning point the gradient = 0 so shape -- horizontal

Immediately past the turning point the gradient is + ve and the shape of the curve is / sloping upwards

ie d/dx(dy/dx) = d²y/dx is +ve (this may be read as "the gradient is becoming more positive at a min. turning point".

Similarly "gradient becomes more negative" at a maximum turning point ie d/dx(dy/dx) is -ve at a max. turning point.