Let `f(x)={{:(3x^2-2x+10, x lt 1),(-2,x gt 1):}`
The set of values of b for which f(x) has greatest value at x=1 is

Options

A. `(-6,-2)`
B. (2,6)
C. `(-6,-2)cup (2,6)`
D. `(-6,6)`

Explanation

To solve the problem, we need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} 3x^2 - 2x + 10 & \text{if } x < 1 \\ -2 & \text{if } x > 1 \end{cases} \]

We want to find the set of values of \( b \) for which \( f(x) \) has a greatest value at \( x = 1 \).

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

First, we need to find the value of \( f(1) \). Since \( f(x) \) is defined piecewise, we need to find the limit of \( f(x) \) as \( x \) approaches 1 from the left:

\[ f(1) = \lim_{x \to 1^-} (3x^2 - 2x + 10) \]

Calculating this limit:

\[ f(1) = 3(1)^2 - 2(1) + 10 = 3 - 2 + 10 = 11 \]

### Step 2: Analyze the behavior of \( f(x) \) around \( x = 1 \)

Next, we need to check the behavior of \( f(x) \) as \( x \) approaches 1 from the left and right.

- For \( x < 1 \), \( f(x) = 3x^2 - 2x + 10 \) is a quadratic function that opens upwards (since the coefficient of \( x^2 \) is positive).

- For \( x > 1 \), \( f(x) = -2 \), which is a constant function.

### Step 3: Check if \( f(x) \) has a maximum at \( x = 1 \)

To ensure that \( f(x) \) has a maximum at \( x = 1 \), we need:

\[ \lim_{x \to 1^-} f(x) > f(1) \quad \text{and} \quad f(1) > f(1^+) \]

Calculating \( f(1^+) \):

\[ f(1^+) = -2 \]

Now we check the inequalities:

1. \( \lim_{x \to 1^-} f(x) = 11 > -2 = f(1^+) \) (True)
2. We need to ensure that \( \lim_{x \to 1^-} f(x) > f(1) \) does not hold since \( f(1) = 11 \).

### Step 4: Find conditions on \( b \)

To find the conditions on \( b \), we need to ensure that the function \( f(x) \) does not exceed \( 11 \) when \( x < 1 \).

We set up the inequality:

\[ 3x^2 - 2x + 10 < 11 \quad \text{for } x < 1 \]

This simplifies to:

\[ 3x^2 - 2x - 1 < 0 \]

Factoring the quadratic:

\[ (3x + 1)(x - 1) < 0 \]

### Step 5: Solve the inequality

To solve the inequality \( (3x + 1)(x - 1) < 0 \), we find the roots:

1. \( 3x + 1 = 0 \Rightarrow x = -\frac{1}{3} \)
2. \( x - 1 = 0 \Rightarrow x = 1 \)

Now we test intervals:

- For \( x < -\frac{1}{3} \): both factors are negative, product is positive.
- For \( -\frac{1}{3} < x < 1 \): one factor is positive, the other is negative, product is negative (satisfies the inequality).
- For \( x > 1 \): both factors are positive, product is positive.

Thus, the solution to the inequality is:

\[ -\frac{1}{3} < x < 1 \]

### Step 6: Determine the set of values for \( b \)

Since we need \( b \) such that \( b^2 - 4 > 0 \) and \( b^2 - 36 < 0 \):

1. \( b^2 - 4 > 0 \Rightarrow b < -2 \text{ or } b > 2 \)
2. \( b^2 - 36 < 0 \Rightarrow -6 < b < 6 \)

Combining these conditions, we find:

- From \( b < -2 \) and \( -6 < b < 6 \), we have \( -6 < b < -2 \).
- From \( b > 2 \) and \( -6 < b < 6 \), we have \( 2 < b < 6 \).

### Final Answer

The set of values of \( b \) for which \( f(x) \) has the greatest value at \( x = 1 \) is:

\[ b \in (-6, -2) \cup (2, 6) \]