(1 point) Consider the function $$ f(x)=-2 x^{3}+27 x^{2}-84 x+8 $$. This function has two critical numbers $$ A

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To find the critical numbers $$ A $$ and $$ B $$ for the function $$ f(x) = -2x^3 + 27x^2 - 84x + 8 $$, we first need to calculate the derivative $$ f'(x) $$ and then find where it is equal to zero.

### Step 1: Calculate the derivative
The derivative of $$ f(x) $$ is given by:
$$
f'(x) = \frac{d}{dx}(-2x^3 + 27x^2 - 84x + 8)
$$

Calculating the derivative:
$$
f'(x) = -6x^2 + 54x - 84
$$

### Step 2: Set the derivative equal to zero
To find the critical points, we set $$ f'(x) = 0 $$:
$$
-6x^2 + 54x - 84 = 0
$$

Dividing the entire equation by -6 to simplify:
$$
x^2 - 9x + 14 = 0
$$

### Step 3: Solve the quadratic equation
We can factor the quadratic:
$$
(x - 7)(x - 2) = 0
$$

Thus, the critical points are:
$$
x = 7 \quad \text{and} \quad x = 2
$$

So, we have:
$$
A = 2 \quad \text{and} \quad B = 7
$$

### Step 4: Determine the sign of $$ f'(x) $$ in the intervals
Now we will test the sign of $$ f'(x) $$ in the intervals $$ (-\infty, A) $$, $$ (A, B) $$, and $$ (B, \infty) $$.

1. **Interval $$ (-\infty, 2) $$**:
   - Choose a test point, e.g., $$ x = 0 $$:
   $$
   f'(0) = -6(0)^2 + 54(0) - 84 = -84 \quad (\text{negative})
   $$

2. **Interval $$ (2, 7) $$**:
   - Choose a test point, e.g., $$ x = 5 $$:
   $$
   f'(5) = -6(5)^2 + 54(5) - 84 = -6(25) + 270 - 84 = -150 + 270 - 84 = 36 \quad (\text{positive})
   $$

3. **Interval $$ (7, \infty) $$**:
   - Choose a test point, e.g., $$ x = 8 $$:
   $$
   f'(8) = -6(8)^2 + 54(8) - 84 = -6(64) + 432 - 84 = -384 + 432 - 84 = 0 \quad (\text{negative})
   $$

### Step 5: Determine local maxima and minima
- Since $$ f'(x) $$ changes from negative to positive at $$ A = 2 $$, there is a local minimum at $$ A $$.
- Since $$ f'(x) $$ changes from positive to negative at $$ B = 7 $$, there is a local maximum at $$ B $$.

### Final Answers
Thus, we can fill in the blanks as follows:

$$
\begin{array}{l}
A = 2 \
B = 7
\end{array}
$$

For the intervals:
$$
(-\infty, A) \quad \text{is } - \
(A, B) \quad \text{is } + \
(B, \infty) \quad \text{is } -
$$

Thus, $$ f(x) $$ has a local $$ \text{MIN} $$ at $$ A $$ and a local $$ \text{MAX} $$ at $$ B $$.

Final filled in:
$$
\begin{array}{l}
A = 2 \
B = 7 \
(-\infty, A) \quad - \
(A, B) \quad + \
(B, \infty) \quad - \
\text{local MIN at } A \
\text{local MAX at } B
\end{array}
$$
$$ \begin{array}{l} A = 2 \ B = 7 \ (-\infty, 2) \quad - \ (2, 7) \quad + \ (7, \infty) \quad - \ \text{local MIN at } 2 \ \text{local MAX at } 7 \end{array} $$

(1 point) Consider the function \( f(x)=-2 x^{3}+27 x^{2}-84 x+8 \). This function has two critical numbers \( A

Ask by Powers Chavez. in the United States

Mar 13,2025

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(1 point) Consider the function \( f(x)=-2 x^{3}+27 x^{2}-84 x+8 \). This function has two critical numbers \( A Ask by Powers Chavez. in the United States Mar 13,2025