Question

	1) A)State the domains (note: [0,inf) for including 0, (0,inf) not include 0) of the following functions (cannot divide by 0 and
		cannot square root negative). B) state which function below is linear? a) g(x)= 1-x b)h(?)=?−1h(x)=√(x^ -1) c)?(?)=−1−?1−?f(x)=-(1-x)/√(1-x) when c does not equal d simplify ?2−?2?−?=(c^2-d^2)/(c-d)= (c^2-d^2)/(c-d)=

Answer 1

A) a) g(x)=1-x, here 1-x can be evealuated for every real number x, so the domain is D= (-inf,inf)

b) h(x)=√(x-1). square root is note defined for negetive numbers. so h(x) is defined only when x-1 is non-negetive, i,e, x-1>=0

x-1>=0 ⇒ x>=1.

So the domain is [1, inf)

c)f(x)=-(1-x)/√(1-x)

a/b is defined only when b ≠ 0, i,e, the dinominator should be non-zero. So f(x) is defined only when √(1-x) ≠ 0   ⇒ x-1 ≠ 0 ⇒ x≠1 .

Also from (b) we have √(x-1) is defined only when x>=1.

So h(x) is deined for x>=1 and x≠1.

The domain is (1,inf)

Function Domain

a) g(x)=1-x (-inf, inf) = R- complete real line

b) h(x)=√(x-1) [1 , inf)

c) (x)=-(1-x)/√(1-x) (1 , inf)

   B) A function f is linear if and only if the following conditions are satisied

1) f(x+y)=f(x)+f(y) and

2) f(cx)=cf(x)

a) g(x)=1-x.

g(1)=0, g(2) = 1, g(1+2)=g(3)=2

g(1)+g(2)=0+1=1

here g(1+2) ≠ g(1)+g(2). So g is not linear

b) h(x)=√(x-1)

h(2)=√(2-1) =√1 = 1

3*h(2) = 3

h(3*2)=h(8) = √7

here h(3*2)  ≠ 3h(2). So h is not linear

c) f(x)=-(1-x)/√(1-x)

f(2)= 1

f(3)= 2/√2 = 1.414

f(2)+f(3)=2.414

f(2+3)=f(5)=4/√4 = 4/2 = 2

here f(2+3) ≠ f(2)+f(3). So f is not linear