^{2}-7

^{2}) and g(x)= √(7-x)

**find a formula for h (x) and the domain of h**( use internal notation)

suppose that f(x) = √(x^{2} -7^{2}) and g(x)= √(7-x)

for each function (h) below,** find a formula for h (x) and the domain of h** ( use internal notation)

a) h(x) = f composed with g of (x)

b) h(x) = g composed with f of (x)

c) h(x)= f composed with f of (x)

d) h(x)= g composed with g of (x)

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For f(x)=√(x²-7²) and g(x)= √(7-x)

a) h(x)=f(g(x))=√(g²(x)-7²)=√(7-x-7²)=√(7-x-49)=√(-x-42)

The domain are those x-values where -x-42≥0 or where x+42≥0 or where x≥-42

For these values h(x)=√(-x-42) takes values in the range [0,∞), or all positive values.

b) h(x)=g(f(x))=√(7-f(x))=√(7-√(x²-7²))=√(7-√(x²-49))

In order to determine the domain we must be able to take square roots so

1. x²-49 must be ≥0 or x²≥49 or either x≤-7 or x≥7 an in addition

√(x²-49) must be ≤7 since we need √(7-√(x²-49)).

√(x²-49)≤7 means that x²-49≤49 or that x²≤98 or that x≤√98=√(2×49)=7√2 or x≥-7√2

Combining the requirements the domain is the two intervals -7√2≤x≤-7 or 7≤x≤7√2

The range of h(x) is h(7) to h(7√2) or from 0 to √7. The function has the same range when the negative boundaries of the domain are used.

c) f(f(x))=√(f²(x)-49)=√(x²-49-49)=√(x²-98). The domain are those values where x²-98≥0 or where

x²≥98 or where x≥√98 or where x≤-√98, That is the two infinite intervals x≤-7√2 and x≥7√2

The range of the function is all the positive numbers.

d) g(g(x))=√(7-g(x))=√(7-√(7-x)). In order to evaluate the square roots we need to have x≤7 for √(7-x) and 7≥√(7-x) for the final square root. This last means that 49≥7-x or that x≥-42. The domain is

-42≤x≤7. The range is from 0 to √7

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