[MATH] average value question

[NEGROWPLEASE]

if you are given a function (say x^3+6x^2+9x+1) how do you find on what intervals or what values the average value of the function is increasing?

do i take the first derivative and set it = 0?

someone help me please.

and by average value i mean the one with 1/(b-a) times the integral of the function.

thakns!

[brachiopod]

Mathematics = plural

=> Abbreviation = Maths (plural)

[gen2]

'Tis many years ago but I seem to remember that the first derivative (dy by dx) will give you the slope (gradient) of the graph.
It will be negative when the function value is decreasing.
It will be positive when the function value is increasing.
It will be zero when the function value is changing between + and - or between - and +. Those will be either maxima, minima, or points of inflection. If you take the first derivative (as you suggest) and equate to zero, you will find these points of change. You will then have to decide on which side of the discovered point the value of the function is increasing, and on which side it is decreasing.

Given
y = x^3 + 6x^2 = 9x + 1
dy/dx = 3x^2 + 12x + 9
........ = 3(x+1)(x+3)

so if 3(x+3)(x+1) = 0
then either (x+3) = 0 or (x+1) = 0
so either x = -3 or x = -1

pick a value less than -3 (lets say -4)
dy/dx = 48 -48 +9 = +9 (slope is positive)
so for values of x < -3 the function value is increasing

pick a value between -1 and -3 (lets say -2)
dy/dx = 12 -24 +9 = -3 (slope is negative)
so for values of -3 < x < -1 the function value is decreasing

Pick a value > -1 (lets say 0)
dy/dx = +9 (slope is positive)
so for values of x > -1 the function value is increasing

I hope this is what you want.

[NEGROWPLEASE]

yeah awesome gen2

thanks a lot