Function f is defined by the equation `f(x)=ax^(2)+(2)/(a)x`. If `f(3)-f(2)=1`, what is the smallest possible value of a?

Options

A. `(1)/(6)`
B. `(1)/(5)`
C. `(1)/(2)`
D. `(1)/(2)`

Answer

To solve the problem, we need to find the smallest possible value of \( a \) given the function \( f(x) = ax^2 + \frac{2}{a}x \) and the condition \( f(3) - f(2) = 1 \).

### Step 1: Calculate \( f(3) \)
We start by substituting \( x = 3 \) into the function:

\[ f(3) = a(3^2) + \frac{2}{a}(3) = 9a + \frac{6}{a} \]

### Step 2: Calculate \( f(2) \)
Next, we substitute \( x = 2 \) into the function:

\[ f(2) = a(2^2) + \frac{2}{a}(2) = 4a + \frac{4}{a} \]

### Step 3: Set up the equation \( f(3) - f(2) = 1 \)
Now, we can set up the equation based on the condition given in the problem:

\[ f(3) - f(2) = \left( 9a + \frac{6}{a} \right) - \left( 4a + \frac{4}{a} \right) = 1 \]

This simplifies to:

\[ (9a - 4a) + \left( \frac{6}{a} - \frac{4}{a} \right) = 1 \]

\[ 5a + \frac{2}{a} = 1 \]

### Step 4: Multiply through by \( a \) to eliminate the fraction
To eliminate the fraction, we multiply the entire equation by \( a \) (assuming \( a \neq 0 \)):

\[ 5a^2 + 2 = a \]

### Step 5: Rearrange the equation
Rearranging gives us a standard quadratic equation:

\[ 5a^2 - a + 2 = 0 \]

### Step 6: Solve the quadratic equation
We can use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5, b = -1, c = 2 \):

\[ b^2 - 4ac = (-1)^2 - 4 \cdot 5 \cdot 2 = 1 - 40 = -39 \]

Since the discriminant is negative, this means there are no real solutions for \( a \).

### Step 7: Re-evaluate the equation
It seems we made a mistake in the setup of the equation. Let's go back to the correct equation from Step 3:

\[ 5a + \frac{2}{a} = 1 \]

Multiply through by \( a \):

\[ 5a^2 + 2 = a \]

Rearranging gives:

\[ 5a^2 - a + 2 = 0 \]

### Step 8: Solve the quadratic equation correctly
Now we need to solve:

\[ 5a^2 - a + 2 = 0 \]

Using the quadratic formula:

\[ a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 5 \cdot 2}}{2 \cdot 5} = \frac{1 \pm \sqrt{1 - 40}}{10} = \frac{1 \pm \sqrt{-39}}{10} \]

Since this also gives us complex solutions, we need to check our calculations.

### Final Step: Check for real solutions
Returning to the original equation \( 5a + \frac{2}{a} = 1 \), we can multiply through by \( a \) again and rearrange:

\[ 5a^2 - a + 2 = 0 \]

Using the quadratic formula correctly, we find:

\[ a = \frac{1 \pm \sqrt{1 - 40}}{10} \]

Since there are no real solutions, we conclude that the smallest possible value of \( a \) is not defined in real numbers.