Given $f(x)=1/8x-3$ and $g(x)=x^3$ , how do you find $(f^-1og^-1)(1)$ ?

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Answer 1 · 2017-02-16T04:44:37

$(f^-1 @ g^-1)(1) = 2root(3)(4)$

To find the inverse functions, switch the $x$ and $y$ values.

Finding $g^-1(x)$

$x = y^3$

$y = root(3)(x)$

Finding $f^-1(x)$

$x = 1/8y - 3$

$x + 3 = 1/8y$

$y = 8x + 24$

Now, the notation $(f @ g)(x)$ is equivalent to $f(g(x))$ .

$(f^-1 @ g^-1)(1) = f^-1(g^-1(1))$

Evaluate now using function notation.

$g^-1(1) = 8(1) + 24 = 8 + 24 = 32$

$f^-1(32/3) = root(3)(32) = root(3)(8 * 4) = 2root(3)(4)$

Hopefully this helps!

Given f(x)=1/8x-3f(x)=1/8x-3 and g(x)=x^3g(x)=x^3, how do you find (f^-1og^-1)(1)(f^-1og^-1)(1)?