Expert: Scotto - 9/3/2009

Question
QUESTION: Please help me find the interval where f(x) - x^3+x^2-5x+2 is increasing and the interval where it is decreasing.

Thank you.

ANSWER: A function is increasing where f'(x)>0 and decreasing where f'(x) < 0.

Looking at f(x) = x^3+x^2-5x+2, f'(x) is 3x^2 + 2x - 5.

This factors into (3x+5)(x-1).

The points where this function is 0 are at -5/2 and -1.

This gives us three intervals.

Lets put in a point in each interval, say, -100, -2, and 100.

At x = -100, both functions are negative, so the answer is positive,
so the function is increasing when x<-5/2.

At x = -1, 3x+5 is 2 and x-1 is -2, so the product is -4, so the result in this interval is negative.  That means the function is decreasing over this interval.

At x = 100, both 3x+5 and x-1 are positive, so the function is increasing again.

At the points x = -5/2 and x = -1, the function is not increasing or decreasing.
These points are called critical points.  x = -5/2 is a local maximum.
x = -1 is a local minimum.



---------- FOLLOW-UP ----------

QUESTION: Thank you for this.  Can you check my work here:
I'm trying to find the indefinite integral for
(integral sign)(25x^4+4x-3)dx

I got
1/6 dx^2 (25x^4+8x-9)

Is that correct?

Answer
I'm not sure what "1/6 dx^2 (25x^4+8x-9)" means.
The term dx^2 usually refers to the 2nd derivative of something.
If we had the function f(x), f"(x) can be rewritten as d^2f/dx^2,
but when dx^2 is by itself, it only arises in advanced calculus and differential equations.


To integrate 25x^4 + 4x - 3, each term needs to be looked at separately.
To integrate, add 1 to the exponet of each term and divide that term by the new exponet.

Since 25x^4 is to the 4th, 4+1=5, so the answer is 25x^5/5 = 5x^5.
Since 4x is to the 1st, 1+1=2, so the answer is 4x^2/2 = 2x^2.
The last term, -3, is a constant.  To integrate, put in an x.  This means the intgral is -3x.
When done with all the terms, a constant C must be added.

This makes the overall integral into 5x^5 + 2x^2 - 3x + C.

Note: A consant K is really Kx^0, so when adding 1 to the exponet, the result is 1.
Dividing by the new exponet, which is 1, doesn't change the answer.  
So the answer is Kx + C.