INCREASING AND DECREASING FUNCTION

                                               Increasing Functions:

The value of Functions increase with increase the value of X.

or

f(x)  increase with increase the value of X-ordinates.

Decreasing Functions:

If the value of function decease with increase in the value of X-coordinate.

here h(x) is deceasing function because the function value decrease with increase in the value of X.

How to find wheather the function is increasing or decreasing within the given interval 

For an increasing function slope of curve is positive within the given interval 

i.e.  f′(x)>0

and

For a deceasing  function slope of curve is negative within the given interval 

i.e.  f′(x)<0

Lets take one example:

f(x)=x²+3x+2

Now calculate the slope i.e. f′(x)=2x+3

for increasing f′(x)>0

hence 

(2x+3)>0

x>(-3/2)

for decreasing function  f′(x)<0

hence

(2x+3)<0

x<(-3/2)

Hence the given function is increasing for x>(-3/2) and decreasing for x<(-3/2) 

Find the range in which function y=sinx-cosx is increasing or decreasing

y′=cosx+sinx

Increasing when y′>0

i.e.

cosx+sinx>0

sinx/cosx>-1

tanx>-1

x<45°

and 

Decreasin when

x>45°

Determine the range in which function y=x³+6x increasing and deceasing 

Given y=x³+6x

y′=3x²+6

since slope y′ is always positive

hence the function is always increasing for any value of x. 

Determine the range in which function y=⟦(x+1)(x-3)⟧³ increasing and deceasing

y=⟦(x+1)(x-3)⟧³

y′=3⟦(x+1)(x-3)⟧²⟦(x+1)+x-3⟧

y′=3(x+1)²(x-3)²(2x-2)

y′=6(x+1)²(x-3)²(x-1)

Rejecting positive term

for incearsing y′>0

or

so only (x-1)>0

i.e.

x>1

so the function is increasing in x>1