Using the method shown in the details, how does this method prove that

[ybhathena]

Homework Statement

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.

Homework Equations

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?

 

[tiny-tim]

hi ybhathena!

(try using the X² button just above the Reply box )

f'(x) = 3x^2 - 6x + 4
= 3(x-1)^2 + 1 > 0
…
… I don't know how by putting it into vertex form it proves that f(x) is always monotone increasing?

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")

 

[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?

 

[Ray Vickson]

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?

Exactly! That is what YOU need to figure out.

 

[ybhathena]

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

 

[Ray Vickson]

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

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.)