# For the given functions f, g, and h, find fog oh, state the exact domain of fog oh. f(x)=eˣ g(x)=logx h(x)=x/x-7PLEASE SHOW ALL WORK!!!!!

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You need to remember how to compose more than two functions, hence:

`(fogoh)(x) = f(g(h(x)))`

Hence, you need to substitute `g(h(x))` for x in equation of function f(x) such that:

`f(g(h(x))) = e^(g(h(x)))`

You need to substitute h(x) for x in equation of function g(x) such that:

`g(h(x)) = log (h(x)) =gt g(h(x)) = log (x/(x-7))`

`` Hence, the equation of the function f(g(h(x))) is: `f(g(h(x))) = e^(log (x/(x-7)))`

You need to remember that the domain of the function comprises the values of x that make the equation of function not to fail.

Hence, you need to consider the condition `x/(x-7) gt 0` or else the logarithm exponent ceases to exist.

You need to solve the inequality `x/(x-7) gt 0` , hence you need to remember that an inequality is positive if both numerator and denominator possess like signs (both positive or both negative).

Considering that both are positive, you need to solve:

`x gt 0 `

`x - 7 gt 0 =gt x gt 7`

Hence, both numerator and denominator are positive if `x in (7,oo).`

`` Considering that both are negative, you need to solve:

`xlt 0` `x - 7 lt 0 =gt x lt 7`

Hence, both numerator and denominator are negative if `x in` `(-oo,0)` .

**Hence, evaluating the domain of the function `f(g(h(x))) = e^(log (x/(x-7)))` yields: `x in` `(-oo,0)U(7,oo).` **