Problem 2The domain and target set of functions f and g is RI The functions are defined as:\item f(æ) = 2x + 3\\\item g(x) = 5x +7|\|\begin(enumerate)[label=(\alph*)]\item fo g?\\\I%Enter your answer below this comment line.\item go fP\\\\%Enter your answer below this comment line.\item (fo g)p\\|%Enter your answer below this comment line.\item f- og-p\\||%Enter your answer below this comment line.\item g-1 o fp\ l|I%Enter your answer below this comment line.\end(enumerate)Are any of the above equal?\\\|%Enter your answer below this comment line.

Name: Problem 2The domain and target set of functions f and g is RI The fun
Uploaded: 2024-03-20T06:38:44.000Z
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Description: Problem 2The domain and target set of functions f and g is RI The functions are defined as:\item f(æ) = 2x + 3\\\item g(x) = 5x +7|\|\begin(enumerate)[label=(\alph*)]\item fo g?\\\I%Enter your answer below this comment line.\item go fP\\\\%Enter you

Given : f (x) = 2 x + 3 and g(x) = 5 x + 7 . Also given that the domain and the range ( target set ) is ℝ, the set of real numbers. To determine : (a) f ∘ g (b) g ∘f(c) f ∘ g -1(d) f-1 ∘ g-1(e) g-1 ∘ f-1 (f) Are any of the above equal? f (x) = 2 x + 3 and g(x) = 5 x + 7 . (a) f ∘ g : f ∘g (x) = f (g(x) )= f(5x+7)= 2(5x+7) + 3 = 10 x +14+3 f ∘g (x) = 10x+17 (b) g ∘ f g ∘f (x) = g (f(x) )= g(2x+3)= 5(2x+3) + 7 = 10 x +15+7 g ∘f (x)= 10x+22 (c) f ∘ g -1 (x ) : To find the inverse of f o g , The inverse is given by, y = 10x+17 interchanging the variable x and y to find the inverse, x = 10 y + 17 10 y = x - 17 y = x10-1710= 110 ( x-17) Therefore, f ∘ g -1 (x ) = x10-1710 (d) f-1 ∘ g-1 : Since, f (x) = 2 x + 3 and g(x) = 5 x + 7 . The inverse of f is given by y = 2 x + 3 , To find the inverse, interchanging the variables, x and y. x = 2 y + 3 2y = x -3y = x2-32 Therefore, the inverse of f (x) is given by f-1 (x) =x2-32. The inverse of g (x) is given by y = 5 x + 7 , To find the inverse, interchanging the variables, x and y. x = 5 y + 7 x = 5 y + 7 5 y = x -7y = x5-75 Therefore, the inverse of g (x) is given by g-1 (x) =x5-75. f-1 ∘ g-1 ( x) = f-1 g-1(x) = f-1 x5-75 = x5-752-32= x10-710-32= x10+-7-1510= x10-2210= 110(x-22) (e) g-1 ∘ f-1 : g-1 ∘ f-1 ( x) = g-1 f-1(x) = g-1 x2-32 = x2-325-75= x10-310-75= x10+-3-1410= x10-1710 = 110 ( x - 17 ) (f) Are any of the above equal : From (c) and (e), it is seen that, f ∘ g -1 (x ) equals g-1 ∘ f-1 . Therefore, f ∘ g -1 = g -1 ∘ f -1 .