How to show that f(x) = x^3 - x^2 + x - 1 is never decreasing

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In summary: But if f'(x) is always positive, the function is never decreasing.Thanks for clarifying! In summary, the given function f(x) = x^3-x^2+x-1 is never decreasing because its derivative, f'(x) = 3x^2 - 2x + 1, is always positive. This can be shown by substituting two consecutive numbers into the derivative and showing that the result is always positive. Alternatively, the function can be shown to be never decreasing by using the theorem that if f'(x) > 0 for all x in an interval, then the function is increasing on that interval.
  • #1
Air
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Homework Statement


Show [itex]f(x) = x^3-x^2+x-1[/itex] is never decreasing.2. The attempt at a solution
[itex]f(x) = x^3-x^2+x-1[/itex]
[itex]f'(x) = 3x^2 - 2x + 1[/itex]
[itex]f''(x) = 6x - 2[/itex]3. The problem that I'm facing
I don't understand what it means by never decreasing. Do I say that when differentiates, it gives a positive quadratic equation hence it never decreasing or when differentiated twice, it forms a positive linear equation hence never decreasing? :confused:
 
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  • #2


For a function to be increasing, for any
[tex]x_1<x_2[/tex];
[tex]f(x_1)<f(x_2)[/tex].
The opposite applies for decreasing functions.
 
  • #3


steven10137 said:
For a function to be increasing, for any
[tex]x_1<x_2[/tex];
[tex]f(x_1)<f(x_2)[/tex].
The opposite applies for decreasing functions.

But, I'm only provided with one function so how can I show:

[itex]f(x_1)<f(x_2)[/itex]

What is [itex]x_1, x_2[/itex]?
 
  • #4


[tex]x_1[/tex] and [tex]x_2[/tex] are simply two consecutive numbers.

It's just logic, take [tex]x_1=2[/tex] and [tex]x_2=3[/tex];
[tex]f(x_1)=5[/tex] and [tex]f(x_2)=20[/tex].

... and you can prove this for any two numbers, aslong as they are consecutive.
 
  • #5


steven10137 said:
[tex]x_1[/tex] and [tex]x_2[/tex] are simply two consecutive numbers.

It's just logic, take [tex]x_1=2[/tex] and [tex]x_2=3[/tex];
[tex]f(x_1)=5[/tex] and [tex]f(x_2)=20[/tex].

... and you can prove this for any two numbers, aslong as they are consecutive.

So, I substitute two consecutive numbers into the function and show that it's increasing.

No differentiation is required for this question?
 
  • #6


Air said:
No differentiation is required for this question?

Indeed. Think about it, if for any two consecutive numbers, their corresponding function value is greater, how can the function be decreasing?

You could also use the theorem that on any interval (a,b), if f '(x) is > 0, then the function is increasing on that range.
Take the interval (1,2) for example; f '(1.5) > 0 and this can be proven for all intervals for this function.

This method is more useful when you are asked questions such as "Find the values of x where the function is increasing or decreasing", as you can use the derivative to find the critical points of the function and evaluate the sign of the derivate over each of the intervals.

Hope this makes sense.
 
  • #7


You have
[tex]f'(x) = 3x^2 - 2x + 1= 3(x^2- (2/3)x+ 1/9)- 1/3+ 1= (x-1/3)^2+ 2/3[/tex]
which is never 0. Checking x= 0, f'(0)= 1> 0 so f' is positive for all x.
 
  • #8


Air said:
I don't understand what it means by never decreasing. Do I say that when differentiates, it gives a positive quadratic equation hence it never decreasing or when differentiated twice, it forms a positive linear equation hence never decreasing? :confused:
If the function never decreases, then it's derivative must be always be positive right? How can you show that is true for f'(x)?
 
  • #9


Air said:
So, I substitute two consecutive numbers into the function and show that it's increasing.

No differentiation is required for this question?

YES. USE DIFFERENTIATION. You've already said the correct thing. f'(x) is a positive quadratic. If the derivative is always positive, a function is increasing. 'never decreasing' would just mean 'always increasing or constant'.
 
  • #10


Thanks. I understand! :smile:
 
  • #11


Strictly speaking, if a function is never decreasing, it's derivative is never negative. That's not the same as saying "always positive" (although in this problem the derivative is always positive).

for example, f(x)= x3 is never decreasing but f'(x)= 3x2 which is NOT always positive: f'(0)= 0.
 

1. How do you show that f(x) is never decreasing?

To show that a function is never decreasing, we need to prove that its derivative is always positive. In this case, we need to take the derivative of f(x) and show that it is always greater than or equal to 0.

2. What is the derivative of f(x) = x^3 - x^2 + x - 1?

The derivative of f(x) = x^3 - x^2 + x - 1 is f'(x) = 3x^2 - 2x + 1. This can be found by using the power rule and the sum rule of differentiation.

3. How can you prove that the derivative of f(x) is always greater than or equal to 0?

We can use the first and second derivative tests to prove that the derivative of f(x) is always greater than or equal to 0. This involves finding the critical points of f(x) and evaluating the sign of the derivative at those points. If the sign is always positive, then the function is never decreasing.

4. Are there any other methods to show that f(x) is never decreasing?

Yes, we can also use the mean value theorem or the definition of a strictly increasing function to show that f(x) is never decreasing. These methods involve using the properties of the function and its derivatives to prove that it is always increasing.

5. Can f(x) ever be decreasing?

No, by definition, a function that is never decreasing means that it is always increasing or constant. Since f(x) = x^3 - x^2 + x - 1 is never decreasing, it can never be decreasing.

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