Show that f and g are inverse functions algebraically.
f(x)=x^3/2
g(x)=3 cuberoot 2x
My work so far:
(3sqrt2x)^3
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2
I need help on the numeral part of the expression
To show that two functions are inverse functions, we need to prove that their compositions result in the identity function. We can do this by showing that f(g(x)) = x and g(f(x)) = x.
Let's start by finding f(g(x)):
f(g(x)) = f(3∛(2x))
= (3∛(2x))^3/2
Next, let's find g(f(x)):
g(f(x)) = g(x^(3/2))
= 3∛(2x^(3/2))
To show that f and g are inverse functions, we need to show that f(g(x)) = x and g(f(x)) = x. Let's start with f(g(x)):
f(g(x)) = (3∛(2x))^3/2
= 3∛(2x)^(3/2)
Now, we need to simplify this expression. To do so, we can use the property of exponents:
2x^(3/2) = 2(x^(1/2))^3
= 2(sqrt(x))^3
= 2(sqrt(x))^2(sqrt(x))
= 2x(sqrt(x))
Substituting this simplification into f(g(x)):
f(g(x)) = 3∛(2x)^(3/2)
= 3∛(2x(sqrt(x)))
= 3(sqrt(2x))
Now, let's find g(f(x)):
g(f(x)) = 3∛(2x^(3/2))
= 3∛(2x(sqrt(x)))
So, we have:
f(g(x)) = 3(sqrt(2x))
g(f(x)) = 3∛(2x(sqrt(x)))
To confirm that f and g are inverse functions, we need to show that f(g(x)) = x and g(f(x)) = x:
f(g(x)) = 3(sqrt(2x)) = x
g(f(x)) = 3∛(2x(sqrt(x))) = x
Therefore, we have shown algebraically that f and g are inverse functions.