Increasing & Decreasing Functions

Find the values of x for which f(x) is an increasing function, given the f(x) equals:

3 + 3x - 3x^2 + x^3

f'(x) = 3 - 6x + 3x^2

Now I've sketched this graph and noticed it has a repeated root at 1 and that is probably why my answer isn't matching up with the answers.
So, Critical values:

(x-1)(3x-3) = 0

x > 1

But this is wrong and the answers show

x \not=1

x \in \mathbb{R}

What does xER mean? :smile:

(edited 8 years ago)

Original post by Psst.

Find the values of x for which f(x) is an increasing function, given the f(x) equals:

3 + 3x - 3x^2 + x^3

f'(x) = 3 - 6x + 3x^2

Now I've sketched this graph and noticed it has a repeated root at 1 and that is probably why my answer isn't matching up with the answers.
So, Critical values:

(x-1)(3x-3) = 0

x > 1

But this is wrong and the answers show

x \not=1

x \in \mathbb{R}

What does xER mean? :smile:

A function

f

is strictly increasing if

f' > 0

.

In your case, you have

f' = 3(x-1)^2

- it has a squared term that will always be positive, no matter what. Except... at

x=-1

we have

f' = 0

which means that the function is stationary there - it's not increasing.

So to summarise we have

f' > 0

as long as

x \neq -1

. Whatever other value of

x

you wish to substitute into

f'

it will always be positive and hence the function will always be increasing.

Reply 3

Psst.

That made perfect sense.

Thank you Dr Zacken :h:

Reply 4

DylanJ42

Original post by Psst.

You need to find the values of x where

3 - 6x + 3x^2

is >0. You worked out that it has a repeated root at x=1, so why don't you sketch

3x^2 - 6x + 3

As for x ∈ R (sorry I cannot latex this), I think it literally translates to x is contained in the Real Numbers

So basically x lies on the number line somewhere, so it includes all the integers, fractions, rational numbers, irrational numbers etc

Reply 5

Simple_student

Original post by Psst.

Find the values of x for which f(x) is an increasing function, given the f(x) equals:

3 + 3x - 3x^2 + x^3

f'(x) = 3 - 6x + 3x^2

Now I've sketched this graph and noticed it has a repeated root at 1 and that is probably why my answer isn't matching up with the answers.
So, Critical values:

(x-1)(3x-3) = 0

x > 1

But this is wrong and the answers show

x \not=1

x \in \mathbb{R}

What does xER mean? :smile:

Hi I've seen this phrase on My C3 course :

XER means x can be any real numbers (pi, e, decimals fractions, mixed numbers, integers, etc..)

Posted from TSR Mobile

Reply 6

Zacken

Original post by DylanJ42

As for x ∈ R (sorry I cannot latex this), I think it literally translates to x is contained in the Real Numbers

\displaystyle x \in \mathbb{R}

Reply 7

Zacken

Original post by Psst.

f'(x) = 3 - 6x + 3x^2

Now I've sketched this graph and noticed it has a repeated root at 1 and that is probably why my answer isn't matching up with the answers.

Original post by DylanJ42

so why don't you sketch

3x^2 - 6x + 3

@OP - you're welcome! :smile:

Reply 8

DylanJ42

Original post by Zacken

\displaystyle x \in \mathbb{R}

Noted, thank you :biggrin:

Reply 9

DylanJ42

Original post by Zacken

@OP - you're welcome! :smile:

didn't read the post properly :facepalm:

Reply 10

Zacken

Original post by DylanJ42

didn't read the post properly :facepalm:

I once wrote a 5 paragraph post only to realise that I missed a condition in the question that rendered all my working obsolete. So I deleted it.

Spoiler

Reply 11

TeeEm

Original post by Psst.

@TeeEm

glad to see you got help as I am teaching until 23.15

Reply 12

DylanJ42

Original post by Zacken

I once wrote a 5 paragraph post only to realise that I missed a condition in the question that rendered all my working obsolete. So I deleted it.

Spoiler

Oh my god how horrible, at that stage I probably would have just posted it anyway and typed at the bottom "whoops didn't see the condition" :eek:

on the bright side though, you wont be forgetting the method to answering whatever the particular question was any time soon

I'd say 5% of all my answers are slightly off just because I read what I want to read :laugh: