To find the relative minimum point of the function [tex]\( f(x) = \frac{x^3}{3} + \frac{5 x^2}{2} + 4 x + 3 \)[/tex], we will follow these steps:
1. Compute the first derivative [tex]\( f'(x) \)[/tex] to find the critical points.
2. Solve [tex]\( f'(x) = 0 \)[/tex] to find the values of [tex]\( x \)[/tex] where the critical points occur.
3. Compute the second derivative [tex]\( f''(x) \)[/tex] to determine the concavity of the function at the critical points.
4. Evaluate [tex]\( f''(x) \)[/tex] at the critical points to identify whether each critical point is a local minimum, maximum, or neither.
5. Determine the corresponding [tex]\( y \)[/tex]-coordinate for the local minimum point.
Let's proceed step-by-step:
### Step 1: Compute the First Derivative
First, compute the first derivative [tex]\( f'(x) \)[/tex] of the given function:
[tex]\[ f'(x) = \frac{d}{dx}\left( \frac{x^3}{3} + \frac{5 x^2}{2} + 4 x + 3 \right) \][/tex]
Using the power rule:
[tex]\[ f'(x) = x^2 + 5x + 4 \][/tex]
### Step 2: Solve [tex]\( f'(x) = 0 \)[/tex]
Set the first derivative equal to zero to find the critical points:
[tex]\[ x^2 + 5x + 4 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 4)(x + 1) = 0 \][/tex]
So, the critical points are:
[tex]\[ x = -4 \quad \text{and} \quad x = -1 \][/tex]
### Step 3: Compute the Second Derivative
Next, compute the second derivative [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}\left( x^2 + 5x + 4 \right) = 2x + 5 \][/tex]
### Step 4: Evaluate [tex]\( f''(x) \)[/tex] at the Critical Points
Evaluate the second derivative at each critical point to determine the nature of each critical point:
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ f''(-4) = 2(-4) + 5 = -8 + 5 = -3 \][/tex]
Since [tex]\( f''(-4) < 0 \)[/tex], the function has a local maximum at [tex]\( x = -4 \)[/tex].
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f''(-1) = 2(-1) + 5 = -2 + 5 = 3 \][/tex]
Since [tex]\( f''(-1) > 0 \)[/tex], the function has a local minimum at [tex]\( x = -1 \)[/tex].
### Step 5: Find the [tex]\( y \)[/tex]-coordinate of the Local Minimum
Evaluate the original function at [tex]\( x = -1 \)[/tex] to determine the corresponding [tex]\( y \)[/tex]-coordinate:
[tex]\[ f(-1) = \frac{(-1)^3}{3} + \frac{5(-1)^2}{2} + 4(-1) + 3 \][/tex]
Simplify the expression:
[tex]\[ f(-1) = \frac{-1}{3} + \frac{5}{2} - 4 + 3 \][/tex]
[tex]\[ f(-1) = -\frac{1}{3} + \frac{5}{2} - 1 \][/tex]
Convert to a common denominator (6):
[tex]\[ f(-1) = -\frac{2}{6} + \frac{15}{6} - \frac{6}{6} \][/tex]
[tex]\[ f(-1) = \frac{7}{6} \][/tex]
Therefore, the local minimum point is:
[tex]\[ \left( -1, \frac{7}{6} \right) \][/tex]
Hence, the correct answer is:
[tex]\[ \text{(B)} \left( -1, \frac{7}{6} \right) \][/tex]
