The range of the function
`f(x)=(e^(x)*logx*5^(x^(2)+2)*(x^(2)-7x+10))/(2x^(2)-11x+12)` is

Options

A. `(-infty,infty)`
B. `[0,infty)`
C. `(3/2,infty)`
D. `(3/2,4)`

Answer

To find the range of the function

\[ f(x) = \frac{e^x \cdot \log x \cdot 5^{x^2 + 2} \cdot (x^2 - 7x + 10)}{2x^2 - 11x + 12}, \]

we will follow these steps:

### Step 1: Factor the numerator and denominator
First, let's simplify the function by factoring both the numerator and the denominator.

- The numerator can be factored as:
\[ e^x \cdot \log x \cdot 5^{x^2 + 2} \cdot (x^2 - 7x + 10) = e^x \cdot \log x \cdot 5^{x^2 + 2} \cdot (x - 5)(x - 2). \]

- The denominator can be factored as:
\[ 2x^2 - 11x + 12 = (2x - 3)(x - 4). \]

Thus, we rewrite the function:
\[ f(x) = \frac{e^x \cdot \log x \cdot 5^{x^2 + 2} \cdot (x - 5)(x - 2)}{(2x - 3)(x - 4)}. \]

### Step 2: Determine the domain
Next, we need to find the domain of the function. The function is defined for all real numbers except where the denominator is zero or where the logarithm is undefined.

- The denominator is zero when:
\[ 2x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2}, \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4. \]

- The logarithm \(\log x\) is undefined for \(x \leq 0\).

Thus, the domain of \(f(x)\) is:
\[ \text{Domain: } x \in \mathbb{R} \setminus \left\{ \frac{3}{2}, 4 \right\} \text{ and } x > 0. \]

### Step 3: Analyze the behavior of the function
Now we need to analyze the behavior of \(f(x)\) as \(x\) approaches the boundaries of the domain and at critical points.

- As \(x \to 0^+\), \(\log x \to -\infty\) and \(f(x) \to 0\).
- As \(x \to \frac{3}{2}^-\) and \(x \to \frac{3}{2}^+\), \(f(x)\) approaches \(\pm \infty\) (the function has a vertical asymptote).
- As \(x \to 4^-\) and \(x \to 4^+\), \(f(x)\) also approaches \(\pm \infty\) (another vertical asymptote).
- As \(x \to \infty\), \(e^x\) and \(5^{x^2 + 2}\) grow much faster than the polynomial terms, leading \(f(x) \to \infty\).

### Step 4: Conclusion about the range
Given that \(f(x)\) approaches \(-\infty\) as \(x\) approaches \(0\) and has vertical asymptotes at \(x = \frac{3}{2}\) and \(x = 4\) where it can take on all values in between, and it also approaches \(+\infty\) as \(x\) goes to infinity, we conclude that:

\[ \text{Range of } f(x) = (-\infty, \infty). \]