Determine where the function is increasing and where it is decreasing?

Question

f(x) = x^3-4x^2+3x

Answer 1

To do this you find the first derivative:
f'(x) = 3x^2 -8x + 3
When f'(x) = 0 there is a miinimum or maximum.
Where is this one increasing to a max and/or min? Set it equal to 0,
3x^2 -8x + 3 = 0; apply the quadratic formula to find the 0's for x:

x = {-(-8) +- sqr[(-8)^2 - 4(3)(3)]}/2(3)
= {8 +- sqr[64-36]}/2 = {8 +- sqr[28]}/2 = {8 +- 2sqr(7)}/6
= (4+-sqr(7))/3 = x1, x2

Now you have two values of x; now look at f(x) in the interval
(-infinity, x1), then (x1,x2), then (x2,+infinity); this will tell you the answer. Make a table showing the values approaching x1,
x1 to x2, then x2 on. And there you have it.

Answer 2

You need to find the maxima & minima:

f(x)dx = 3x² - 8x + 3 = 0
x = (8 ± √(64 - 36)) / 6 = 4/3 ± (√7)/3

So, the function is increasing when x < 4/3 - (√7)/3, decreasing between 4/3 - (√7)/3 < x < 4/3 + (√7)/3, and increasing when x > 4/3 + (√7)/3

Answer 3

Increasing/decreasing is found from the first derivative.
f'(x) = 3x^2 - 8x + 3
Set this equal to zero and solve for x, then test the intervals to find out whether they are positive or negative.
If the interval is positive, then f(x) is increasing on that interval. If it is negative, then f(x) is decreasing on that interval.