Find the range of the following functions.
`f(x)=log_(e)(3x^(2)-4x+5)`

Answer

To find the range of the function \( f(x) = \log_e(3x^2 - 4x + 5) \), we will follow these steps:

### Step 1: Determine when the function is defined
The function \( f(x) \) is defined when the argument of the logarithm is positive:
\[ 3x^2 - 4x + 5 > 0 \]

### Step 2: Analyze the quadratic expression
To analyze the quadratic expression \( 3x^2 - 4x + 5 \), we can calculate its discriminant \( D \):
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot 5 = 16 - 60 = -44 \]
Since the discriminant is negative, the quadratic does not have any real roots and is always positive (as the leading coefficient is positive).

### Step 3: Conclude the domain
Since \( 3x^2 - 4x + 5 > 0 \) for all \( x \in \mathbb{R} \), the function \( f(x) \) is defined for all real numbers:
\[ \text{Domain of } f(x) = (-\infty, \infty) \]

### Step 4: Set up the equation for the range
Let \( y = f(x) = \log_e(3x^2 - 4x + 5) \). To find the range, we can rewrite this in exponential form:
\[ e^y = 3x^2 - 4x + 5 \]

### Step 5: Rearrange to form a quadratic equation
Rearranging gives us:
\[ 3x^2 - 4x + (5 - e^y) = 0 \]
This is a quadratic equation in \( x \).

### Step 6: Use the discriminant condition
For \( x \) to be real, the discriminant of this quadratic must be non-negative:
\[ D = (-4)^2 - 4 \cdot 3 \cdot (5 - e^y) \geq 0 \]
Calculating the discriminant:
\[ 16 - 12(5 - e^y) \geq 0 \]
\[ 16 - 60 + 12e^y \geq 0 \]
\[ 12e^y \geq 44 \]
\[ e^y \geq \frac{44}{12} = \frac{11}{3} \]

### Step 7: Solve for \( y \)
Taking the natural logarithm of both sides:
\[ y \geq \log_e\left(\frac{11}{3}\right) \]

### Step 8: State the range
Since \( y \) can take any value greater than or equal to \( \log_e\left(\frac{11}{3}\right) \), the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \left[ \log_e\left(\frac{11}{3}\right), \infty \right) \]