C1 Proofs - Show that f(x) is increasing or decreasing.....

awlright

I reli don't understand this concept and i'm confused on how the examiner will give me marks. Could somebody please explain this concept please...

For example...
f(x) = 2 − x − x3.
Show that f(x) is decreasing for all values of x. (4 marks)

Show that there isn't a turning point. i.e. differentiate and equate to zero. There shouldn't be a limit. Then say something like the coefficients of the x terms are negative which shows it's decreasing.

Reply 2

Differentiate the equation, and you will find that the first derivative is always negative (with all values of x)

The first derivative gives you the gradient of the line, and if the gradient is negative for all values of x then the line is always decreasing

Reply 3

OP

So i say f '(x) = 1 - 3x^2
At SP f'(x) = 0
Thus, well, i'm stuck now.......

Reply 4

awlright

So i say f '(x) = 1 - 3x^2
At SP f'(x) = 0
Thus, well, i'm stuck now.......

-1-3x^2

This is negative for all values of x

Reply 5

username219321

0

14

you need to prove f'(x) is always < 0

Reply 6

Student#1

matt2k8

you need to prove f'(x) is always < 0

Uh?!

f''(x) should always be negative.. Why is everyone saying f'(x) should be negative!?

Edit: sorry i thought you were talking about maximum and minimum..

Reply 7

OP

That's where i'm stuck, (sorry for differentiating wrong), even from there, how can you assume -1-3x^2<0 or -1-3x^2 is less than or equal to -1

Reply 8

Student#1

Uh?!

f''(x) should always be negative.. Why is everyone saying f'(x) should be negative!?

Edit: sorry i thought you were talking about maximum and minimum..

lol

Reply 9

Khodu

19

Is this Edexcel or OCR? :unsure:

Reply 10

Rosh_123

10

It's edexcel I think, unless the OCR sylabus changed since last year.

Reply 11

awlright

That's where i'm stuck, (sorry for differentiating wrong), even from there, how can you assume -1-3x^2<0 or -1-3x^2 is less than or equal to -1

You don't assume anything --> 3x^2 is always positive

Therefore -1-3x^2 is always negative (i.e. a negative number minus any positive number is always negative, and if 3x^2 is always positive then f'(x) is always negative)

Reply 12

nathan00

awlright

That's where i'm stuck, (sorry for differentiating wrong), even from there, how can you assume -1-3x^2<0 or -1-3x^2 is less than or equal to -1

well -1 is negative and -3x^2 is negative... so -1-3x^2 is negative and hence less than zero

Reply 13

Khodu

Is this Edexcel or OCR? :unsure:

Its definitely edexcel but it could also be OCR

Reply 14

OP

So how should i express the first derivative:

from -1-3x^2=f'(x) what do i say next, i don't understand the mark scheme answer:

f ′(x) = −1 − 3x2
x2 ≥ 0 for all real x ⇒ −1 − 3x2 ≤ −1 (i especially don't understand how this bit has been done...)
∴ f ′(x) < 0 ⇒ f(x) is decreasing for all values of x (4)

Reply 15

OP

And the exam board is OCR

Reply 16

Benny_b

8

Adhavan

Its definitely edexcel but it could also be OCR

well then....not definitely edexcel is it??

Reply 17