FUNCTIONS

• State the domain and range of the following relation. Is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

The domain is all x -values.

and

Range is all y-values.

Hence in the above sets

Domain {2,4,3,6,2}

and

Range {-3,6,-1,6,3}

for  function {(2,1),(10,5)}

hence 
Domain {2,10}
and
Range {1,5}

Function and Relations

Determine the Domain of the function:

y=1/(x²+3x+2)

y is defined for all value of x except

where (x²+3x+2)=0

x²+2x+x+2=0

from this x=-2 and x=-1

hence the range is all real value of x except -1 and -2

Determine the Domain of the function:

y=x/√(x²-3x+2)

y is defined for
where (x²-3x+2)>0  here we take greater than zero(>)                    because the square root is in the denominator. If equal to zero then the (x²-3x+2)=0 and y=∞, which is not possible.                                               

x²-2x-x+2>0

x(x-2)-1(x-2)>0

(x-1)(x-2)>0

from this we conclude that

x<1 and x>2

hence the range of the function is x<1 and x>2

Determine the Domain of the function:

y=√(x²-5x+6)

y is defined for
where (x²-5x+6)≥0       

here we take greater than equal to zero(≥)                                           because the square root is in the numerator                                                         . 

x²-3x-2x+6≥0

x(x-3)-2(x-3)≥0

(x-3)(x-2)≥0

from this we conclude that

x≤2 and x≥3

Determine the range of the function:

y=x²+4

Range is the all y value.

To find the range write x in term of x

i.e.

y-4=x²

or

x=√(y-4)

for x to be defined (y-4)≥0

or

y≥0

hence the range of the function is y≥0

Determine the range of the function:

y-4=4/x²

Range is the all y value.

for range:

write the x in terms of y

i.e.

x²=4/(y-4)

or

x=2/√(y-4)

for x to be defined (y-4)>0

or

y>4

hence the range of function is y>4

Determine the domain and range of the y=9/(√(x²-4))

for domain

for y to be defined x²-4>0

from this

(x-2)(x+2)>0

or

x>2 and x<-2

hence the domain of the function x>2 and x<-2

for range

for x to be defined

write x in terms of y

i.e. 

x²-4=(81/y²)

or

x²=(81/y²)+4

or

x=√((81/y²)+4)

x is defined for all value of y except when y=0

hence range is all real value of y except when y=0

What is the domain and range of the function f(x)=1/x+3
Given function

f(x)=1/x+3

Domain is the all x value for which f(x) is defined, hence the f(x) is defined for all real number except at x=0

Domain= for all real number except x=0

Range

All value of y for which x is defined

To find range write x in terms of y

y=1/x+3

y-3=1/x

x=1/(y-3)

hence x is defined for all real number except when y-3=0 i.e. y=3

Range: All real number except y=3

If f(x)=2x+2 , then what is f(f(3)) equal to?

f(x)=2x+3

now

f(f(x))=2f(x)+3

Now

f(f(3))=2f(3)+3

but 

f(3)=2×3+2=8

f(f(x))=2f(x)+3=2×8+3

f(f(x))=19

What is the domain and range of √9+x^2?

y=√9+x²

Domain is all value of x for y is defined.

9+x²≥0 which is true for all value of x.

hence domain is all real value.

and 

Range

y²=9+x²

or x=√y²-9

range is all value of y for which x is defined

y²-9≥0

or (y-3)(y+3)≥0

or

range is⟦-∞,-3⟧U⟦3,∞⟧

What is the maximum value of f(x) =√3 sin x + cos x?
Max value of asinx+bcosx is √(a²+b²)

so max value of √3sinx+cosx is √(√3)²+1²=√(3+1)=√4=2

•   

$\mathbf{\color{green}{\mathit{y} = \dfrac{\mathit{x}^2 + \mathit{x} - 2}{\mathit{x}^2 - \mathit{x} - 2}}}$