Increasing and Decreasing functions

[andreipanait]

My book says the following:

In order to find wether a function is increasing or decreasing we must first find the derivative of the function. In this case, x^3-4x^2 + 3x turns into 3x^2 - 8x + 3.

So far this makes sense. But then:

Now to find out at which points the function is increasing or decreasing we must first find the roots.

Unfortunately I plugged in the derived form of the equation (3x^2 - 8x + 3) and you get a negative in the square root. Thereby you cannot find the root.

Is there another way of going about it or am I being silly and missed something :S
-Thank you for your help

 

[andreipanait]

:S

Oh did someone re-structure my post :S If so thank you

 

[stapel]

Hint: Look at the graph of f(x) = x³ - 4x² + 3x. Does it have any "turns"? Or is it always increasing? If the former, then your derivative should have real roots. If the latter, then not.

I don't know what you mean by "plugging in" the derivative. What x-value did you plug into it? Or did you mean something else? (Since the discriminant -- the bit inside the square root -- is (-8)² - 4(3)(3) = 64 - 36, there is no negative inside the root in the Quadratic Formula. So you must be referring to something else, is why I ask.)

Note: If the derivative has no zeroes (or undefined spots), then the original function is always either increasing or decreasing. These are perfectly valid possibilities. For instance, f'(x) = x² + 4 has no (real) zeroes, and the graph of f(x) = (1/3)x³ + 4x has no turning points.

Eliz.

 

[andreipanait]

OHHH

Hint: Look at the graph of f(x) = x³ - 4x² + 3x. Does it have any "turns"? Or is it always increasing? If the former, then your derivative should have real roots. If the latter, then not.

I don't know what you mean by "plugging in" the derivative. What x-value did you plug into it? Or did you mean something else? (Since the discriminant -- the bit inside the square root -- is (-8)² - 4(3)(3) = 64 - 36, there is no negative inside the root in the Quadratic Formula. So you must be referring to something else, is why I ask.)

Note: If the derivative has no zeroes (or undefined spots), then the original function is always either increasing or decreasing. These are perfectly valid possibilities. For instance, f'(x) = x² + 4 has no (real) zeroes, and the graph of f(x) = (1/3)x³ + 4x has no turning points.

Eliz.

My dearest friend, I am a moron lol
I see now my mistake, for some reason i forgot that that there is a b^2 inside the square root of the quadratic formula. I thank thee for ur help, now it makes sense
Thanks again!!