• anonymous
If f(x) = (√x-8) and g(x) =x+4, find the function h(x) such that h(x)= (f o g)^-1. Be sure to give a domain restriction if necessary. Need to know how to get the answer as well because I have to show work. Really lost. Help?
  • katieb
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  • anonymous
-If an equation is a function of x it means that when you sub in a value for x you'll get another value, the function value, 'out' of the function. -When an equation is a function of another function, this means that the inner function will take on values as x changes, and this in turn will affect the outer function. take the example above, you have: h(x) - h is a function that varies as x changes g(x) - g is a function that varies as x changes f(x) - f is a function that varies as x changes (f o g) denotes that you first have to sub an x value into g, get the result and sub that into the variable for the function f. Let's break down your question: \[(f \space o\space g)=f(g(x))=f(x+4)\] so now where ever you see an x in the f(x) function you'll need to replace it with [x+4] \[(f\space o\space g) = f(x+4) = \sqrt{x+4}-8\]
  • anonymous
now you have a compound function, let's introduce the inverse and consider restrictions: \[(f\space o\space g)^{-1}=\frac{1}{\sqrt{x+4}-8}\] restrictions state that you can't have values less than zero under a root sign, so: \[x + 4 \ge 0\] \[x\ge -4\] But we also can't ever divide by zero so the denominator must be greater than zero: \[\sqrt{x+4}-8\ne0\] \[\sqrt{x+4} \neq 8\] \[x+4 \neq 64\] \[x\neq 60\] since the domain must adhere to both these restrictions you must form a union with them: \[\left\{ x \in \mathbb{R} |x \ge-4, x\neq 60\right\}\]

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