Range of `f(x) = ln(3x^2-4x+5)` is

Options

A. `[ln (11/3),oo)`
B. `[ln 10, oo)`
C. `[ln (11/6),oo)`
D. `[ln (11/12),oo)`

Explanation

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

### Step 1: Analyze the function inside the logarithm
First, we need to analyze the quadratic function \( g(x) = 3x^2 - 4x + 5 \).

### Step 2: Determine the nature of the quadratic function
The coefficient of \( x^2 \) in \( g(x) \) is positive (3), which means the parabola opens upwards. Therefore, it will have a minimum value.

### Step 3: Find the minimum value using the vertex formula
The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Here, \( a = 3 \) and \( b = -4 \).

\[ x = -\frac{-4}{2 \cdot 3} = \frac{4}{6} = \frac{2}{3} \]

### Step 4: Calculate the minimum value of \( g(x) \)
Now, substitute \( x = \frac{2}{3} \) back into \( g(x) \) to find the minimum value.

\[ g\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 5 \]
\[ = 3 \cdot \frac{4}{9} - \frac{8}{3} + 5 \]
\[ = \frac{4}{3} - \frac{8}{3} + 5 \]
\[ = -\frac{4}{3} + 5 = -\frac{4}{3} + \frac{15}{3} = \frac{11}{3} \]

### Step 5: Determine the range of \( g(x) \)
Since \( g(x) \) is a quadratic function that opens upwards, the range of \( g(x) \) is:

\[ \left[\frac{11}{3}, \infty\right) \]

### Step 6: Find the range of \( f(x) \)
Now, since \( f(x) = \ln(g(x)) \), we need to find the range of \( f(x) \) based on the range of \( g(x) \).

The natural logarithm function \( \ln(x) \) is defined for \( x > 0 \) and is increasing. Therefore, we can find the range of \( f(x) \) by evaluating \( \ln \) at the minimum value of \( g(x) \).

\[ \text{Minimum value of } g(x) = \frac{11}{3} \]
\[ \text{Thus, } f(x) \text{ will have a minimum value of } \ln\left(\frac{11}{3}\right) \]

### Step 7: Write the final range of \( f(x) \)
Since \( g(x) \) approaches infinity, \( f(x) \) will also approach infinity. Therefore, the range of \( f(x) \) is:

\[ \left[\ln\left(\frac{11}{3}\right), \infty\right) \]

### Final Answer
The range of \( f(x) = \ln(3x^2 - 4x + 5) \) is:

\[ \left[\ln\left(\frac{11}{3}\right), \infty\right) \]

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