Find the value of f(1) that the function `f(x)= (9(x^(2/3)-2x^(1/3)+1))/((x-1)^(2)), x ne 1` is continuous at x =1

Answer

To find the value of \( f(1) \) such that the function

\[ f(x) = \frac{9(x^{2/3} - 2x^{1/3} + 1)}{(x-1)^2}, \quad x \neq 1 \]

is continuous at \( x = 1 \), we need to ensure that

\[ \lim_{x \to 1} f(x) = f(1). \]

### Step 1: Calculate the limit as \( x \) approaches 1

We start by simplifying the expression for \( f(x) \):

\[ f(x) = \frac{9(x^{2/3} - 2x^{1/3} + 1)}{(x-1)^2}. \]

### Step 2: Factor the numerator

We recognize that the expression \( x^{2/3} - 2x^{1/3} + 1 \) can be factored. We can rewrite it as:

\[ x^{2/3} - 2x^{1/3} + 1 = (x^{1/3} - 1)^2. \]

Thus, we can substitute this back into our function:

\[ f(x) = \frac{9((x^{1/3} - 1)^2)}{(x - 1)^2}. \]

### Step 3: Relate \( (x-1) \) to \( (x^{1/3} - 1) \)

We can express \( x - 1 \) in terms of \( x^{1/3} - 1 \) using the identity for the difference of cubes:

\[ x - 1 = (x^{1/3})^3 - 1^3 = (x^{1/3} - 1)((x^{1/3})^2 + (x^{1/3}) + 1). \]

### Step 4: Substitute back into the function

Thus, we can rewrite \( (x - 1)^2 \) as:

\[ (x - 1)^2 = \left((x^{1/3} - 1)((x^{1/3})^2 + (x^{1/3}) + 1)\right)^2 = (x^{1/3} - 1)^2 \cdot ((x^{1/3})^2 + (x^{1/3}) + 1)^2. \]

### Step 5: Simplify \( f(x) \)

Now substituting this back into \( f(x) \):

\[ f(x) = \frac{9(x^{1/3} - 1)^2}{(x^{1/3} - 1)^2 \cdot ((x^{1/3})^2 + (x^{1/3}) + 1)^2} = \frac{9}{((x^{1/3})^2 + (x^{1/3}) + 1)^2}. \]

### Step 6: Evaluate the limit as \( x \) approaches 1

Now we can find the limit as \( x \to 1 \):

\[ \lim_{x \to 1} f(x) = \frac{9}{((1^{1/3})^2 + (1^{1/3}) + 1)^2} = \frac{9}{(1^2 + 1 + 1)^2} = \frac{9}{3^2} = \frac{9}{9} = 1. \]

### Step 7: Set \( f(1) \)

Since we want the function to be continuous at \( x = 1 \), we set:

\[ f(1) = \lim_{x \to 1} f(x) = 1. \]

Thus, the value of \( f(1) \) that makes the function continuous at \( x = 1 \) is

\[ \boxed{1}. \]