^{3}and then exchanges the roles of x and y, i.e.

^{3}and solve for y

^{1/3}= 2y - 5 => x

^{1/3}+ 5 = 2y and y = (1/2)(x

^{1/3}+ 5)

are functions g(h(x)) and h(g(x)) inverses?

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First, I suggest that you graph the two equations given.

2 functions which are inverses will be symmetric with respect to the line y = x.

Second to compute the inverse of g(x), write y = (2x-5)^{3} and then exchanges the roles of x and y, i.e.

x = (2y-5)^{3} and solve for y

x^{1/3} = 2y - 5 => x^{1/3} + 5 = 2y and y = (1/2)(x^{1/3} + 5)

Now add this function to your graph and look at the relation between this function and g(x)!

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## Comments

PROVIDED WHAT YOU MEAN for h(x) is CUBE ROOT OF (x + 5/2).Because in that case, g(f(x)) = x ... that's the criterion for mutual inverses.Many students are not able to express, in words, a radical having 3 in its index (little number in the radical symbol's 'notch'.)