Prove that g(x)=7x^3+2 and g^-1(x)=cube root(x-2)/7 are inverses of each other using the compositions

Answer 1

The inverse composition rule says that Equation 1 evaluated at the value of Equation 2 should equal Equation 2 evaluated at the value of Equation 1. Another way to express this is that g(g^-1(x)) should equal g^-1(g(x)).

Let's start with g(g^-1(x)). This will equal the g(x) equation with every x value replaced with the g^-1(x) equation.

g(x) = 7x³ + 2

g(g^-1(x)) = 7(((x - 2) / 7)^1/3)³ + 2

First, let's deal with the exponents. Since one exponent is being raised to the other, we will multiple the exponents. (1/3) x 3 = 1. Anything raised to the power of 1 is itself (8¹ = 8), so this basically gets rid of the exponents. So:

g(g^-1(x)) = 7((x - 2) / 7) + 2

Next, the 7 in the numerator will cancel with the 7 in the denominator, so we're left with:

g(g^-1(x)) = x - 2 + 2

Finally, the 2 and -2 will combine to equal 0.

g(g^-1(x)) = x

There isn't any more simplifying we can do, so we're done with this one.

Now we have to check that g^-1(g(x)) also simplifies to x.

g^-1(x) = ((x-2) / 7)^1/3

g^-1(g(x)) = (((7x³ + 2 - 2) / 7)^1/3

First, we can combine the 2 and -2 in the innermost parentheses to equal 0.

g^-1(g(x)) = (7x³ / 7)^1/3

Next, the 7 in the numerator will cancel with the 7 in the denominator.

g^-1(g(x)) = (x³)^1/3

Finally, we'll multiply the exponents together to find our new exponent. 3 x (1/3) = 1, and x¹ = x, so:

g^-1(g(x)) = x

Since both g(g^-1(x)) and g^-1(g(x)) simply to x, we have proven through composition that they are inverse functions.

I hope this helps!

Answer 2

g(x) and g^-1(x) are inverses if their composition g(g^-1(x)) = x. Let's try it:

g(x) = 7x³ + 2

g(g^-1(x)) = 7(g^-1(x))³ + 2

g(g^-1(x)) = 7(³√((x-2)/7))³ + 2

g(g^-1(x)) = 7((x-2)/7) + 2

Can you finish it? It should equal x.