The range of the function f(x) = frac{e^x ln | vedclass

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(A) First,factorize the numerator and denominator: $f(x) = \frac{e^x \ln x \cdot 5^{(x^2 + 2)}(x - 2)(x - 5)}{(2x - 3)(x - 4)}$ Note that the function

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$( - \infty, \infty )$

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(A) First,factorize the numerator and denominator: $f(x) = \frac{e^x \ln x \cdot 5^{(x^2 + 2)}(x - 2)(x - 5)}{(2x - 3)(x - 4)}$ Note that the function is defined for $x > 0$ due to $\ln x$ and $x 
eq \frac{3}{2}, x 
eq 4$. In the interval $\left( \frac{3}{2}, 4 ight)$,the function is continuous. Evaluating the limits at the boundaries: $\lim_{x 	o (3/2)^+} f(x) = \frac{(+) \cdot \ln(1.5) \cdot (+) \cdot (1.5 - 2)(1.5 - 5)}{(+) \cdot (0^+)} = \frac{(+) \cdot (+) \cdot (+) \cdot (-) \cdot (-)}{(+) \cdot (0^+)} = \frac{+}{0^+} = \infty$ Wait,let us re-evaluate the sign: At $x 	o (3/2)^+$,the denominator $(2x-3)$ is $(0^+)$ and $(x-4)$ is negative. Numerator: $e^x$ is $(+)$,$\ln x$ is $(+)$,$5^{(x^2+2)}$ is $(+)$,$(x-2)$ is negative,$(x-5)$ is negative. So,$\frac{(+) \cdot (+) \cdot (+) \cdot (-) \cdot (-)}{(+) \cdot (-)} = \frac{+}{-} = -\infty$. At $x 	o 4^-$,the denominator $(2x-3)$ is $(+)$,$(x-4)$ is $(0^-)$. Numerator: $e^x$ is $(+)$,$\ln x$ is $(+)$,$5^{(x^2+2)}$ is $(+)$,$(x-2)$ is $(+)$,$(x-5)$ is negative. So,$\frac{(+) \cdot (+) \cdot (+) \cdot (+) \cdot (-)}{(+) \cdot (0^-)} = \frac{-}{-} = \infty$. Since the function is continuous on $\left( \frac{3}{2}, 4 ight)$ and goes from $-\infty$ to $\infty$,the range is $(-\infty, \infty)$.

The range of the function $f(x) = \frac{e^x \ln x \cdot 5^{(x^2 + 2)}(x^2 - 7x + 10)}{2x^2 - 11x + 12}$ is

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