Range of the function f(x) = 3 + 2^x + 4^x is | vedclass

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(A) Given the function $f(x) = 3 + 2^x + 4^x$. Since $2^x > 0$ for all $x \in \mathbb{R}$,it follows that $4^x = (2^x)^2 > 0$. Let $t = 2^x$,where $t

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$(3, \infty)$

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(A) Given the function $f(x) = 3 + 2^x + 4^x$. Since $2^x > 0$ for all $x \in \mathbb{R}$,it follows that $4^x = (2^x)^2 > 0$. Let $t = 2^x$,where $t \in (0, \infty)$. Then the function becomes $g(t) = 3 + t + t^2$. Since $t > 0$,the minimum value of $t + t^2$ approaches $0$ as $t 	o 0^+$. As $t 	o \infty$,$t + t^2 	o \infty$. Therefore,the range of $t + t^2$ for $t > 0$ is $(0, \infty)$. Adding $3$ to this range,the range of $f(x)$ is $(3 + 0, 3 + \infty) = (3, \infty)$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

Range of the function $f(x) = 3 + 2^x + 4^x$ is

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