second derivative?

Question

CAn someone explain 'how we determine the maximum and minimum points using the second derivative of a function?
I wanna know how the whole thing works..like how the negative value of the 2nd derivative gives the maximum value.
Pleeaassee help me out with this!!

Answer 1

The second derivative is the rate of change of the first derivative, which is the instantaneous slope. So the second derivative tells you how fast the slope is changing - think of it as the slope of the slope.

If you think of it like the slope of the slope, then a negative value of the second derivative means the first derivative is decreasing, and a positive value means teh first derivative is increasing.

If you've got that, then this part is simple. If you have a function, f, and at some point x0 the first derivative of f goes to zero (df/dx|x=x0 = 0), then you are at a local max or min. So you take the second derivative and evaluate it at x = x0. If the second derivative is positive, the first derivative is increasing. If you are at the local max, the moving in either direction in x causes the value of f(x) to decrease, hence the slope decreases (the second derivative is negative). If you are at a local min, then moving in either direction in x cause the value of f(x) to increase, hence the slope increases (the second derivative is positive).

So think of sitting at either the local min or the local max. What ahppens to the slope of the function as you move away? If it increases, the second derivative is increasing and you are at a local min. If it decreases, the second derivative is decreasing and you are at a local max.

Answer 2

f'(x) = unfavourable for all x the only factor it quite is reducing for all values may well be a function it quite is self sufficient of any fee. this means that the 1st spinoff is a relentless, ok, the place ok < 0. f''(x) = ? The spinoff of a relentless is 0. for this reason the 2d spinoff of f''(x) is 0. Graph: combine f'(x) = okay to verify f(x) INTEG(f'(x) = ok) = f(x) = kx, the place ok< 0 So the graph of this function is a diagonal line going from north and west of 0 to south and east of 0 and passing throughout the beginning (0,0). The slope relies upon upon ok

Answer 3

graphically is the best way to explain it