Question 1190841: 3. If f(x)=square root 2x^2-1 and g(x)=x^1/2, find (and simplify)
a) (f+g)(x)
b) (f-g)(x)
c) (f*g)(x)
d) (f/g)(x)
e) (g/f)(x)
f) (f*g)(x)=f(g(x))
g) (g*f)(x)=g(f(x))
h) Determine whether g and f are inverse functions.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here are the solutions:
**a) (f+g)(x) = f(x) + g(x)**
(f+g)(x) = √(2x² - 1) + x^(1/2)
**b) (f-g)(x) = f(x) - g(x)**
(f-g)(x) = √(2x² - 1) - x^(1/2)
**c) (f*g)(x) = f(x) * g(x)**
(f*g)(x) = √(2x² - 1) * x^(1/2) = x^(1/2)√(2x² - 1) = √(x(2x² - 1)) = √(2x³ - x)
**d) (f/g)(x) = f(x) / g(x)**
(f/g)(x) = √(2x² - 1) / x^(1/2) = √(2x² - 1)/√x = √((2x² - 1)/x) = √(2x - 1/x)
**e) (g/f)(x) = g(x) / f(x)**
(g/f)(x) = x^(1/2) / √(2x² - 1) = √x/√(2x² - 1) = √(x/(2x² - 1))
**f) (f*g)(x) = f(g(x))**
(f*g)(x) = f(x^(1/2)) = √(2(x^(1/2))² - 1) = √(2x - 1)
**g) (g*f)(x) = g(f(x))**
(g*f)(x) = g(√(2x² - 1)) = (√(2x² - 1))^(1/2) = (2x² - 1)^(1/4)
**h) Are g and f inverse functions?**
No, they are not inverse functions. If they were, then (f*g)(x) and (g*f)(x) would both simplify to just x. As we can see from parts (f) and (g), this is not the case. Also, the domains of f(x) and g(x) need to be considered. For f(x) to be defined, 2x²-1 must be greater or equal to zero, so x must be greater or equal to 1/√2 or less than or equal to -1/√2. g(x) is only defined for x greater or equal to 0. For inverse functions, the range of f(x) must equal the domain of g(x) and vice versa. These functions do not satisfy this condition.
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