The range of the real valued function f(x) = | vedclass

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(A) Let $g(x) = 5 + 4x - x^2$. We can rewrite this as $g(x) = -(x^2 - 4x - 5) = -(x^2 - 4x + 4 - 9) = 9 - (x - 2)^2$. Since $(x - 2)^2 \geq 0$,the max

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$(-\infty, 2]$

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(A) Let $g(x) = 5 + 4x - x^2$. We can rewrite this as $g(x) = -(x^2 - 4x - 5) = -(x^2 - 4x + 4 - 9) = 9 - (x - 2)^2$. Since $(x - 2)^2 \geq 0$,the maximum value of $g(x)$ is $9$ (at $x = 2$). Also,for the logarithm to be defined,we require $g(x) > 0$,so $9 - (x - 2)^2 > 0$,which implies $(x - 2)^2 As $x$ varies in $(-1, 5)$,$g(x)$ takes all values in the interval $(0, 9]$. Since $f(x) = \log_3(g(x))$,the range of $f(x)$ is $(\log_3(0^+), \log_3(9)]$. As $g(x) 	o 0^+$,$\log_3(g(x)) 	o -\infty$. As $g(x) 	o 9$,$\log_3(g(x)) 	o \log_3(9) = 2$. Therefore,the range is $(-\infty, 2]$.

The range of the real valued function $f(x) = \log_3(5 + 4x - x^2)$ is

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