If `x in R`, then the expression `9^(x) - 3^(x) + 1` assumes

Options

A. all real values
B. all real values greater than 0
C. all real values greater than 3/4
D. all real values greater than 1/4

Answer

To solve the expression \( f(x) = 9^x - 3^x + 1 \) and find its range, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ f(x) = 9^x - 3^x + 1 \]
We can rewrite \( 9^x \) as \( (3^x)^2 \):
\[ f(x) = (3^x)^2 - 3^x + 1 \]

### Step 2: Substitute \( y = 3^x \)
Let \( y = 3^x \). Then, we can rewrite the expression in terms of \( y \):
\[ f(x) = y^2 - y + 1 \]

### Step 3: Analyze the quadratic expression
The expression \( y^2 - y + 1 \) is a quadratic function in \( y \). To find its minimum value, we can complete the square:
\[ f(y) = y^2 - y + 1 = \left( y - \frac{1}{2} \right)^2 - \frac{1}{4} + 1 \]
This simplifies to:
\[ f(y) = \left( y - \frac{1}{2} \right)^2 + \frac{3}{4} \]

### Step 4: Determine the minimum value
The term \( \left( y - \frac{1}{2} \right)^2 \) is always non-negative (i.e., \( \geq 0 \)). Therefore, the minimum value of \( f(y) \) occurs when \( y - \frac{1}{2} = 0 \), which gives:
\[ f(y) \geq \frac{3}{4} \]

### Step 5: Conclusion about the range
Since \( y = 3^x \) can take any positive value (as \( x \) varies over all real numbers), the expression \( f(y) \) can take values starting from \( \frac{3}{4} \) and going to infinity. Thus, the range of the expression \( f(x) \) is:
\[ f(x) \geq \frac{3}{4} \]

### Final Answer
The expression \( 9^x - 3^x + 1 \) assumes values in the range:
\[ \left[ \frac{3}{4}, \infty \right) \]

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