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Verified answer
Jawba.
range f(x) = x² + 4x - 5 --> f(x) ∈ rii
range s(x)= √(x² + 2x - 3) --> s(x) ≥ 0
b. f(x). g(x) = (x² + 4x - 5)(3x-2) = 3x³ + 10x² - 23x + 10
c. fog(x) = f (g(x)) = f(3x-2)= (3x-2)² + 4(3x-2) - 5
= 9x² -12x - 9
d. gof (x) = g{ f(x)} = g(x² + 4x - 5)
= 3(x² + 4x - 5) - 2
= 3x² + 12x - 17
e. goh(x) = g { h(x)} = g{(3x -2)/(2x+3)}
= 3 (3x -2)/(2x + 3 ) -1
= (9x - 6)/(2x + 3) - 1
= (9x -6 -2x- 3)/(2x + 3)
= (7x - 9)/(2x + 3)
f. fofo g (x) = f { fog(x)} = f{3x² +12 x -17)
= (3x² +12x -17)² + 4(3x² +12x - 17) - 5
= 9x⁴ + 72x³ + 54x² -360x + 216
g. fogoh(-1) = fog {h(-1)} = fog{ (3(-1)-2)/(2(-1)+3)} = f(-5/1)= f(-5)
= (-5) +4(-5) - 5
= 25 - 20 -5
= 0
h. hogof(1) = ho g{ f(1} = hog {1+4 -5)
= hog(0) = h { g(0)} = h (3(0) - 2)
= h (-2)
= (3(2) + 2)/ (2(-2) + 3
= 8/-1
= - 8