The function f(x) is defined by `f(x)={log_[4x-3](x^2-2x+5) `if 3/4 < x< 1 & x>1
4 when x=1}

Options

A. is continuous at x=1
B. is discontinous at x==1 since `f(1^(+))` does not exist though `f(1^(-))` exsits
C. is disccontinous at x=1 since `f(1^(-))` odes not exsits though `f(1^(+))` exsits
D. is discontinuous at x=1 since neither `f(1^(+)) nor f^(1^(-))` exists.

Answer

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \log_{4x-3}(x^2 - 2x + 5) & \text{if } \frac{3}{4} < x < 1 \text{ or } x > 1 \\ 4 & \text{if } x = 1 \end{cases} \]

we need to check the continuity at the point \( x = 1 \). For \( f(x) \) to be continuous at \( x = 1 \), the following condition must hold:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

### Step 1: Calculate \( f(1) \)

From the definition of the function, we have:

\[ f(1) = 4 \]

### Step 2: Calculate \( \lim_{x \to 1^-} f(x) \)

To find \( \lim_{x \to 1^-} f(x) \), we use the expression for \( f(x) \) when \( x < 1 \):

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \log_{4x-3}(x^2 - 2x + 5) \]

Using the change of base formula for logarithms, we can rewrite this as:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{\log(x^2 - 2x + 5)}{\log(4x - 3)} \]

### Step 3: Evaluate \( x^2 - 2x + 5 \) at \( x = 1 \)

Calculating \( x^2 - 2x + 5 \):

\[ 1^2 - 2(1) + 5 = 1 - 2 + 5 = 4 \]

### Step 4: Evaluate \( 4x - 3 \) at \( x = 1 \)

Calculating \( 4x - 3 \):

\[ 4(1) - 3 = 4 - 3 = 1 \]

### Step 5: Substitute back into the limit

Now substituting these values into the limit:

\[ \lim_{x \to 1^-} f(x) = \frac{\log(4)}{\log(1)} = \frac{\log(4)}{0} \]

Since \( \log(1) = 0 \), this limit approaches infinity:

\[ \lim_{x \to 1^-} f(x) = \infty \]

### Step 6: Calculate \( \lim_{x \to 1^+} f(x) \)

Next, we calculate \( \lim_{x \to 1^+} f(x) \):

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \log_{4x-3}(x^2 - 2x + 5) \]

Using the same steps as above:

\[ \lim_{x \to 1^+} f(x) = \frac{\log(4)}{\log(1)} = \frac{\log(4)}{0} \]

This limit also approaches infinity:

\[ \lim_{x \to 1^+} f(x) = \infty \]

### Step 7: Conclusion

Now we have:

\[ \lim_{x \to 1^-} f(x) = \infty, \quad \lim_{x \to 1^+} f(x) = \infty, \quad f(1) = 4 \]

Since \( \lim_{x \to 1^-} f(x) \) and \( \lim_{x \to 1^+} f(x) \) do not equal \( f(1) \), the function \( f(x) \) is discontinuous at \( x = 1 \).

### Final Answer

The function \( f(x) \) is discontinuous at \( x = 1 \).

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