Find functions $$ f $$ and $$ g $$ so that $$ f \circ g=H $$. $$ H(x)=\sqrt{x^{2}+6} $$

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The functions $$ f $$ and $$ g $$ are $$ f(y) = \sqrt{y} $$ and $$ g(x) = x^2 + 6 $$. To find functions $$ f $$ and $$ g $$ such that $$ f \circ g = H $$, where $$ H(x) = \sqrt{x^2 + 6} $$, we can think about how to decompose the function $$ H $$ into two simpler functions. Let's start by defining $$ g(x) $$ as the inner function that will transform $$ x $$ into something that $$ f $$ can then process. A natural choice is to let $$ g(x) $$ be the expression inside the square root: $$ g(x) = x^2 + 6. $$ Next, we need to define $$ f $$ such that it takes the output of $$ g(x) $$ and produces $$ H(x) $$. Since $$ H(x) = \sqrt{g(x)} $$, we can define $$ f $$ as follows: $$ f(y) = \sqrt{y}, $$ where $$ y $$ is the output of $$ g(x) $$. Now we can check if $$ f \circ g = H $$: $$ f \circ g(x) = f(g(x)) = f(x^2 + 6) = \sqrt{x^2 + 6}. $$ This matches our original function $$ H(x) $$. Thus, we have found the functions: $$ g(x) = x^2 + 6, $$ $$ f(y) = \sqrt{y}. $$ In conclusion, the functions $$ f $$ and $$ g $$ such that $$ f \circ g = H $$ are: $$ f(y) = \sqrt{y}, \quad g(x) = x^2 + 6. $$

Find functions \( f \) and \( g \) so that \( f \circ g=H \). \( H(x)=\sqrt{x^{2}+6} \)

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Find functions \( f \) and \( g \) so that \( f \circ g=H \). \( H(x)=\sqrt{x^{2}+6} \)

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