Stitz-Zeager_College_Algebra_e-book

We see that the points 1 1 and 2 1 are both elements

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Unformatted text preview: (h ◦ g )(x) = 2−√x+3 +1 . To ( ) Hence, (h ◦ g )(x) = simplification: domain of h ◦ g , we look to the step before any keep the square root happy, we require x + 3 ≥ 0 √ √ or x ≥ −3. Setting the denominator equal to zero gives 2 − x + 3 + 1 = 0 or x + 3 = 3. Squaring both sides gives us x + 3 = 9, or x = 6. Since x = 6 checks in the original equation, √ 2 − x + 3 + 1 = 0, we know x = 6 is the only zero of the denominator. Hence, the domain of h ◦ g is [−3, 6) ∪ (6, ∞). 5. To find (h ◦ h)(x), we substitute the function h into itself, h(h(x)). • inside out : We insert the expression h(x) into h to get (h ◦ h)(x) = h(h(x)) = h 2x x+1 2x x+1 2x +1 x+1 4x x + 1 · (x + 1) 2x (x + 1) +1 x+1 2 = = 284 Further Topics in Functions 4x · (x + 1) x+1 = 2x · (x + 1) + 1 · (x + 1) x+1 4x $ · (x$$ + 1) X $ $ 1$ (x + $$ 1) 2x $ + 1) · (x$$ + x + 1 X$ $$ 1 (x + $$ 1) 4x 3x + 1 = = • outside in : This approach yields (h ◦ h)(x) = h(h(x)) = 2(h(x)) h(x) + 1 2x x+1 2x +1 x+1 4x 3x + 1 2 = = same algebra as...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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