Answer :
To determine where the function \( f(x) = \frac{1}{2} x^2 + 5x + 6 \) is increasing, we need to analyze its derivative.
1. Find the derivative of \( f(x) \):
[tex]\[ f(x) = \frac{1}{2} x^2 + 5x + 6 \][/tex]
The derivative of \( f(x) \), denoted as \( f'(x) \), is found by differentiating \( f(x) \) with respect to \( x \):
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{2} x^2 + 5x + 6 \right) \][/tex]
Applying standard differentiation rules:
[tex]\[ f'(x) = \frac{1}{2} \cdot 2x + 5 = x + 5 \][/tex]
2. Find the critical points by setting the derivative to zero:
[tex]\[ f'(x) = x + 5 = 0 \][/tex]
Solving for \( x \):
[tex]\[ x = -5 \][/tex]
This tells us that \( x = -5 \) is a critical point of the function and indicates a potential minimum or maximum.
3. Analyze where the function is increasing or decreasing:
Since the derivative \( f'(x) = x + 5 \) changes sign around the critical point \( x = -5 \), we can determine the intervals by testing the sign of \( f'(x) \) on either side of \( x = -5 \).
- For \( x < -5 \), choose \( x = -6 \):
[tex]\[ f'(-6) = -6 + 5 = -1 \][/tex]
Since \( -1 \) is negative, \( f(x) \) is decreasing for \( x < -5 \).
- For \( x > -5 \), choose \( x = 0 \):
[tex]\[ f'(0) = 0 + 5 = 5 \][/tex]
Since \( 5 \) is positive, \( f(x) \) is increasing for \( x > -5 \).
4. Determine the interval of increase:
Based on the analysis, \( f(x) \) is increasing for \( x > -5 \).
Therefore, the interval over which the graph of \( f(x) \) is increasing is \((-5, \infty)\).
So, the correct answer is:
[tex]\[ \boxed{(-5, \infty)} \][/tex]
1. Find the derivative of \( f(x) \):
[tex]\[ f(x) = \frac{1}{2} x^2 + 5x + 6 \][/tex]
The derivative of \( f(x) \), denoted as \( f'(x) \), is found by differentiating \( f(x) \) with respect to \( x \):
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{2} x^2 + 5x + 6 \right) \][/tex]
Applying standard differentiation rules:
[tex]\[ f'(x) = \frac{1}{2} \cdot 2x + 5 = x + 5 \][/tex]
2. Find the critical points by setting the derivative to zero:
[tex]\[ f'(x) = x + 5 = 0 \][/tex]
Solving for \( x \):
[tex]\[ x = -5 \][/tex]
This tells us that \( x = -5 \) is a critical point of the function and indicates a potential minimum or maximum.
3. Analyze where the function is increasing or decreasing:
Since the derivative \( f'(x) = x + 5 \) changes sign around the critical point \( x = -5 \), we can determine the intervals by testing the sign of \( f'(x) \) on either side of \( x = -5 \).
- For \( x < -5 \), choose \( x = -6 \):
[tex]\[ f'(-6) = -6 + 5 = -1 \][/tex]
Since \( -1 \) is negative, \( f(x) \) is decreasing for \( x < -5 \).
- For \( x > -5 \), choose \( x = 0 \):
[tex]\[ f'(0) = 0 + 5 = 5 \][/tex]
Since \( 5 \) is positive, \( f(x) \) is increasing for \( x > -5 \).
4. Determine the interval of increase:
Based on the analysis, \( f(x) \) is increasing for \( x > -5 \).
Therefore, the interval over which the graph of \( f(x) \) is increasing is \((-5, \infty)\).
So, the correct answer is:
[tex]\[ \boxed{(-5, \infty)} \][/tex]
