Find the domain and the range of the square root function, f : R + ∪ { 0 } → R f ( x ) = x for all non-negative real numbers.

Given, f x = x we know that function is not defined if expression under even root is negative. So, f x will be defined if x ≥ 0 Therefore, the domain of function is D f = [ 0 , ∞ ) . Let f x = y = x ∵ x ≥ 0 ⇒ y ≥ 0 Therefore, the range of function is R f = [ 0 , ∞ ) .

Find the domain and the range of the square root function, f : R + ∪ { 0 } → R f ( x ) = x for all non-negative real numbers.

Given, f x = x we know that function is not defined if expression under even root is negative. So, f x will be defined if x ≥ 0 Therefore, the domain of function is D f = [ 0 , ∞ ) . Let f x = y = x ∵ x ≥ 0 ⇒ y ≥ 0 Therefore, the range of function is R f = [ 0 , ∞ ) .

Find the domain and the range of the cube root function, f : R → R : f ( x ) = x 1 3 for all x ∈ R

Given f ( x ) = x 1 3 ∀ x ∈ R Here f ( x ) = x 1 3 Because all real numbers have a cube root. There is no value of which makes the function undefined. The domain of the above function would be, Domain ( f ) = R Consider f ( x ) = y Then y = x 1 3 ⇒ y 3 = x Since f ( x ) is continuous, it follows that for every value of y there would be a cube of that value. Range ( f ) = R

Let f : R → R : f ( x ) = x + 1 and g : R → R : g ( x ) = 2 x - 3 . Find ( f + g ) ( x )

Given f ( x ) = x + 1 and g ( x ) = 2 x - 3 T o f i n d : ( f + g ) ( x ) ( f + g ) ( x ) = f ( x ) + g ( x ) = ( x + 1 ) + ( 2 x - 3 ) = 3 x - 2 Therefore, ( f + g ) ( x ) = 3 x - 2

Let f : R → R : f ( x ) = x + 1 and g : R → R : g ( x ) = 2 x - 3 . Find ( f - g ) ( x )

Given f x = x + 1 and g x = 2 x - 3 We know that ( f - g ) ( x ) = f ( x ) - g ( x ) = ( x + 1 ) - ( 2 x - 3 ) = 4 - x Therefore, ( f - g ) ( x ) = 4 - x

Let f : R → R : f ( x ) = x + 1 and g : R → R : g ( x ) = 2 x - 3 . Find ( f g ) ( x )

Given f ( x ) = x + 1 and g ( x ) = 2 x - 3 We know that ( f g ) ( x ) = f ( x ) · g ( x ) = ( x + 1 ) ( 2 x - 3 ) = 2 x 2 - 3 x + 2 x - 3 = 2 x 2 - x - 3 Therefore. ( f g ) ( x ) = 2 x 2 - x - 3

Let f : R → R : f ( x ) = x + 1 and g : R → R : g ( x ) = 2 x - 3 . Find f g x

Given f ( x ) = x + 1 and g ( x ) = 2 x - 3 Consider, f g ( x ) = f ( x ) g ( x ) ⇒ f g ( x ) = x + 1 2 x - 3 Therefore, f g ( x ) = x + 1 2 x - 3

Let f : R → R : f ( x ) = 2 x + 5 and g : R → R : g ( x ) = x 2 + x . Find ( f + g ) ( x )

Given f ( x ) = 2 x + 5 and g ( x ) = x 2 + x We know that ( f + g ) ( x ) = f ( x ) + g ( x ) = ( 2 x + 5 ) + x 2 + x = x 2 + 3 x + 5 Therefore, ( f + g ) ( x ) = x 2 + 3 x + 5

Let f : R → R : f ( x ) = 2 x + 5 and g : R → R : g ( x ) = x 2 + x . Find ( f - g ) ( x )

Given f ( x ) = 2 x + 5 and g ( x ) = x 2 + x To find: ( f - g ) ( x ) We know that ( f - g ) ( x ) = f ( x ) - g ( x ) = ( 2 x + 5 ) - x 2 + x = - x 2 + x + 5 Therefore, ( f - g ) ( x ) = - x 2 + x + 5

Let f : R → R : f ( x ) = 2 x + 5 and g : R → R : g ( x ) = x 2 + x . Find ( f g ) ( x )

Given f ( x ) = 2 x + 5 and g ( x ) = x 2 + x We know that ( f g ) ( x ) = f ( x ) · g ( x ) = ( 2 x + 5 ) · x 2 + x = 2 x 3 + 2 x 2 + 5 x 2 + 5 x Therefore, ( f g ) ( x ) = 2 x 3 + 7 x 2 + 5 x