Both fractions have the same denominator so it's going to be easy. First set up your equation:
Add the whole numbers:
Now add the fractions:
add both together:
is an improper fraction so change it:
Since 6 is one more than 5, add 1 to the whole number and subtract the numerator and denominator(6 - 5 = 1) and make the remaining the new numerator. That leaves you with
Your answer is
The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.
Domain is the values of x in the function represented by y=f(x), for which y exists.
THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".
Now we assume the and
So here since g(x) is a polynomial function so it exists for all real x.
<em> </em>does not exists when
, so the domain of f(x) is given by all real x except 6.
Now,
So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.
But domain of both f(x) and g(x) consists of the value x=4.
Hence the statement is not TRUE universarily.
Thus the given statement about the composition of function is FALSE.
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