f(x)=root (x-2),g(x)=x^2+x.the function (f*g)(x) is defined for x = all real numbers except for interval a,b 1.calculate the value of a and b 2.find the range o

Question

f(x)=root (x-2),g(x)=x^2+x.the function (f*g)(x) is defined for x = all real numbers except for interval a,b

1.calculate the value of a and b

2.find the range of (f.g)

Answer 1

f(x) = √(x - 2)

 

g(x) = x² + x

 

 

 

The domain is the set of x values in which a function is defined.

 

For f(x), we cannot have a negative number under the square-root.  So the domain is all real numbers greater than or equal to 2.

 

In interval notation:  [2, ∞)

 

 

g(x) is a parabola.  A parabola has a domain of all real numbers.

                                           

Because f(x) has a restriction, (f*g)(x) is defined for all real numbers except for interval -∞, 1.99.

 

a = -∞

b = 1.99  (the digit 9 repeats)

 

 

To find the range, we evaluate f(x) as x approaches infinity, starting from x=2.  The range is the limit of f(x).

 

f(2) = √(2 - 2)

       = 0

 

f(20) = √(20 - 2)

        = √18

       

f(100) = √(100 - 2)

          = √98

 

f(500) = √(500 - 2)

          = √498

 

f(2000) = √(2000 - 2)

            = √1998

 

f(10000) = √(10000 - 2)

              = √9998

 

As you can see, as x approaches infinity, the value of f(x) does not reach a constant value.  Therefore, the limit does not exist, and the range of f(x) is in the interval [0, ∞).

 

g(x) does have a range.  By putting g(x) in vertex form, the vertex will give us the range.  g(x) in vertex form is

 

g(x) = (x² + x + 1/4) - 1/4

 

g(x) = (x + 1/2)(x + 1/2) - 1/4

 

g(x) = (x + 1/2)² - 1/4

 

 

The vertex of g(x) is (-1/2, -1/4).  The y-coordinate of the vertex tells us the range.   Since the parabola opens upward, the range of g(x) is in the interval (-1/4, ∞).

 

 

Therefore, the range of (f*g)(x) in interval notation is

 

[0, ∞)