The functions f and g are defined as follows. f(x)= 1/(x+2)

g(x)= square root of: (x-1)

(a)
State which value of x cannot be included in the domain of f.

(b)
Find the inverse of g(x)

(c) Calculate fg(10)

x != -2, x^2 + 1, 1/5

Step-by-step explanation:

(a) It can be understood that nothing can divide 0

thus denominator cannot be 0 meaning x cannot be -2.

(-2 + 2) = 0

(b) y = sqrt(x-1)

inverse meaning we make x the subject

square both sides

y^2 = x - 1

y^2 + 1 = x

inverse = x^2 + 1

(c) composite function:

pass g(10) as domain of f

g(10) = 3, f(3) = 1/(5) = 1/5

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81