# Homework Help: Using the method shown in the details, how does this method prove that

1. Mar 16, 2013

### ybhathena

1. The problem statement, all variables and given/known data

Solution:

Let f(x) = x^3 - 3x^2 + 4x + 1
f'(x) = 3x^2 - 6x + 4
= 3(x-1)^2 + 1 > 0

Therefore f(x) is always monotone increasing.
From f(0) = 1,
x> 0 and f(x) > 1

and therefore proves the inequality.

2. Relevant equations

3. The attempt at a solution

I understand how they got the derivative and how they got it into vertex form, but I don't know how by putting it into vertex form it proves that f(x) is always monotone increasing? Can someone please help me understand this method?

2. Mar 16, 2013

### tiny-tim

hi ybhathena!

(try using the X2 button just above the Reply box )
the fact that it's the vertex (as you call it) is irrelevant

all that matters is that 3(x-1)2 + 1 is a square plus 1,

so it must always be ≥ 1, and so it's always > 0

(this is called "completing the square")

3. Mar 16, 2013

### ybhathena

But what do you mean by it is always positive. Isnt it, the slope of the tangent line ? So then what would it mean that the slope is always positive?

4. Mar 16, 2013

### Ray Vickson

Exactly! That is what YOU need to figure out.

5. Mar 16, 2013

### ybhathena

Oh I think I get it, if the slope is always increasing that means the function is never negative hence monotone increasing right?

6. Mar 16, 2013

### Ray Vickson

All you need is that the slope is > 0; it does not matter if the slope is increasing, decreasing, or constant. Many strictly increasing functions have slopes that are decreasing; in such cases the slopes are always > 0 but are numerically smaller as x is increasing. (That means that for larger x the function increases at a smaller rate, but it still increases.)