Answer :
To determine the interval over which the function [tex]\( f(x) = \frac{1}{2}x^2 + 5x + 6 \)[/tex] is increasing, follow these steps:
1. Find the First Derivative:
The first derivative of the function [tex]\( f(x) \)[/tex] will help us understand where the function is increasing or decreasing. The first derivative, [tex]\( f'(x) \)[/tex], represents the slope of the function at any point [tex]\( x \)[/tex].
[tex]\[ f(x) = \frac{1}{2} x^2 + 5 x + 6 \][/tex]
Taking the derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{2} x^2 + 5x + 6 \right) = x + 5 \][/tex]
2. Find the Critical Points:
The critical points occur where the first derivative is zero or undefined. In this case, we set the first derivative to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ f'(x) = x + 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -5 \][/tex]
3. Determine the Behavior Around the Critical Point:
To determine where the function is increasing, we need to analyze the sign of the first derivative before and after the critical point [tex]\( x = -5 \)[/tex].
- For [tex]\( x < -5 \)[/tex], choose a test point, say [tex]\( x = -6 \)[/tex]:
[tex]\[ f'(-6) = -6 + 5 = -1 \quad (\text{negative value}) \][/tex]
- For [tex]\( x > -5 \)[/tex], choose a test point, say [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = 0 + 5 = 5 \quad (\text{positive value}) \][/tex]
The derivative [tex]\( f'(x) \)[/tex] is negative for [tex]\( x < -5 \)[/tex], indicating the function is decreasing in this interval. The derivative [tex]\( f'(x) \)[/tex] is positive for [tex]\( x > -5 \)[/tex], indicating the function is increasing in this interval.
4. Conclusion:
Based on our analysis, we conclude that the function [tex]\( f(x) = \frac{1}{2} x^2 + 5 x + 6 \)[/tex] is increasing in the interval where the first derivative is positive, which is:
[tex]\[ (-5, \infty) \][/tex]
Therefore, the correct interval over which the graph of [tex]\( f(x) = \frac{1}{2} x^2 + 5 x + 6 \)[/tex] is increasing is [tex]\(\boxed{(-5, \infty)}\)[/tex].
1. Find the First Derivative:
The first derivative of the function [tex]\( f(x) \)[/tex] will help us understand where the function is increasing or decreasing. The first derivative, [tex]\( f'(x) \)[/tex], represents the slope of the function at any point [tex]\( x \)[/tex].
[tex]\[ f(x) = \frac{1}{2} x^2 + 5 x + 6 \][/tex]
Taking the derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{2} x^2 + 5x + 6 \right) = x + 5 \][/tex]
2. Find the Critical Points:
The critical points occur where the first derivative is zero or undefined. In this case, we set the first derivative to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ f'(x) = x + 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -5 \][/tex]
3. Determine the Behavior Around the Critical Point:
To determine where the function is increasing, we need to analyze the sign of the first derivative before and after the critical point [tex]\( x = -5 \)[/tex].
- For [tex]\( x < -5 \)[/tex], choose a test point, say [tex]\( x = -6 \)[/tex]:
[tex]\[ f'(-6) = -6 + 5 = -1 \quad (\text{negative value}) \][/tex]
- For [tex]\( x > -5 \)[/tex], choose a test point, say [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = 0 + 5 = 5 \quad (\text{positive value}) \][/tex]
The derivative [tex]\( f'(x) \)[/tex] is negative for [tex]\( x < -5 \)[/tex], indicating the function is decreasing in this interval. The derivative [tex]\( f'(x) \)[/tex] is positive for [tex]\( x > -5 \)[/tex], indicating the function is increasing in this interval.
4. Conclusion:
Based on our analysis, we conclude that the function [tex]\( f(x) = \frac{1}{2} x^2 + 5 x + 6 \)[/tex] is increasing in the interval where the first derivative is positive, which is:
[tex]\[ (-5, \infty) \][/tex]
Therefore, the correct interval over which the graph of [tex]\( f(x) = \frac{1}{2} x^2 + 5 x + 6 \)[/tex] is increasing is [tex]\(\boxed{(-5, \infty)}\)[/tex].
