Consider the function fx Squareroot x 3 Find the domain an

Consider the function f(x) = Squareroot x + 3 Find the domain and range of f. Domain: Range: Calculate the values of f at x = 1, f(1) = at x = 1/4, f(1/4) = at x = -1, f(-1) = at x = 4, f(4) = Is f one-to-one (injective)? onto (surjective)? bijective? Explain your answers.

Solution

Given,

f(x)=sqrt(x)+3

a) Domain : [0,infinite),{x|x>=0}

(Set the function inside the square root >=0 and get the interval i.e x>=0)

Range : [3,infinite),{y|y>=3}

(let y=f(x)=sqrt(x)+3. For different values of x get the y values in a tabular form and draw the graph & obtain the range value)

b) f(1)=sqrt(1)+3=4

   f(1/4)=sqrt(1/4)+3=1/2+3=7/2

f(-1)=sqrt(-1)+3=i+3 (since, -1=i²)

   f(4)=sqrt(4)+3=5

c) Given function is ;

f(x)=sqrt(x)+3

from the b bit where ,

f(1)=4

f(-1)=i+3

f(4)=5

f(-4)=2i+3

Hence, for each value of x, f(x) has its one d=different values.

Therefore it exhibits the properties of one to one i.e injective function.