Question

A function is defined by the equation y=(2x)/(x+1). Show that it is increasing for all real values of x except for x≠-1.

How do you show?
Does anyone know the standard working/presentation for this sort of question?

Answer 1

to show increasing or decreasing we use differentiation

Answer 2

we can show by slope or double derivative

Answer 3

derivative is most easy

Answer 4

find dy/dx

Answer 5

and equate it to zero

Answer 6

now u will get some value of x

Answer 7

Oh double derivative. Thanks^^

what's the slope method? is it the -1 0 1 \—/ one?

Answer 8

now find second derivative

Answer 9

if it comes <0 for that value of x than it is inc

Answer 10

double derivative is just differenciating twice, i know haha :3

Answer 11

yeah ...but at first derivative find x value for which derivative =0
and put that value in second derivative than

Answer 12

calculus is pretty :)
can you think of any other way ?

Answer 13

what's the slope method? is it the -1 0 1 \—/ one?

Answer 14

just show that the slope is positive

Answer 15

actually i don't think its supposed to be by the dy^2 method because its not even covered in the topic i'm working on yet.

Answer 16

\[\large \dfrac{dy}{dx} \gt 0 \implies \text{y is increasing}\]

Answer 17

first derivative will do

Answer 18

so what does the 'except for x≠-1' part mean?

Answer 19

the function was not defined at x = -1
so don't bother about it

Answer 20

ohh okay

i've got a quick question though, if you don't mind answering that :3

find the range(s) of values of x for which the following function is increasing:
y=2x^3-3x^2+6

I differenciated then did the > 0 thing
ended up with
6x^2-6x>0
6x(x-1)>0
6x>0, x-1>0
x>0, x>1

but x>0 is wrong
apparently the answer is x<0
why? D:

Answer 21

6x(x-1) > 0

Answer 22

gives you two cases :

case 1 :
x > 0 AND x-1 > 0

case 2 :
x < 0 AND x-1 < 0

Answer 23

work each case separately

Answer 24

case 1 :
x > 0 AND x-1 > 0

x > 0 AND x > 1
the strictest is `x > 1`

Answer 25

whoa why are there two cases? there isn't a square root or anything

Answer 26

consider this :

ab > 0

Answer 27

for what values of `a` and `b` is that product is positive ?

Answer 28

when both `a` and `b` are positive, the product `ab` is positive, right ?

Answer 29

yep

Answer 30

is that the only case ?

Answer 31

both negative

Answer 32

yes the product is positive when both `a` and `b` are negative too

Answer 33

thats exactly what we are doing in the problem above

Answer 34

6x(x-1) > 0

divide 6 both sides

x(x-1) > 0

think of it as product of two numbers : `x` and `x-1`

Answer 35

so i would end up getting
x>0, (x-1)>0, -x>0, -(x-1)>0

i'll reject the third one
and i solve for the rest?

Answer 36

careful about using AND and OR

Answer 37

AND is intersection
OR is union

Answer 38

I don't think I've encountered those terms used that way..

Answer 39

you would end up getting

x>0 AND (x-1)>0

OR

-x>0 AND -(x-1)>0

Answer 40

ahhh okay yes i have

Answer 41

its okay, they are just what you would expect them to be

Answer 42

so
x>0 and x>1

or

x>0 and x>1

?

Answer 43

when you multiply a negative number, the inequality flips

Answer 44

so
x>0 and x>1

or

x<0 and x<1

Answer 45

oh shoot yes

x>0 and x>1 or x<0 and x>1

okay so why should i choose the second one?

Answer 46

looks you have a typo, check once