To check over which interval the graph of our function is increasing, we are going to use the derivative test. The derivative test uses the derivative of the function to locate its critical points, and then check over which intervals the derivative of the function is positive or negative. If the derivative of the function is positive over an interval, then the function is increasing over that interval, but if the derivative of the function is negative over an interval, then the function is decreasing over that interval.
The first thing we are going to do is take the derivative of our function
[tex] f(x)=x^2+5x+6 [/tex]
[tex] f'(x)=\frac{dy}{dx} x^2+\frac{dy}{dx}5x+\frac{dy}{dx}6 [/tex]
[tex] f'(x)=2x+5 [/tex]
Next, we are going to set the derivative equal to zero and solve for [tex] x [/tex] to find the critical points of our function
[tex] 2x+5=0 [/tex]
[tex] 2x=-5 [/tex]
[tex] x=-\frac{5}{2} [/tex]
Our function only has only critical point [tex] x=-\frac{5}{2} [/tex]. That critical point divides the domain of our function in tow intervals: (-∞,-2.5) and (-2.5,∞)
Now we are going to take test values in each interval and evaluate the derivative for those test values
- For the interval (-∞,-2.5), we are going to use the test value -2.6
[tex] f'(x)=2x+5 [/tex]
[tex] f'(-2.6)=2(-2.6)+5 [/tex]
[tex] f'(-2.6)=-5.2+5 [/tex]
[tex] f'(-2.6)=--0.2 [/tex]
Since the derivative is negative over interval (-∞,-2.5), the function [tex] f(x)=x^2+5x+6 [/tex] is decreasing over the interval (-∞,-2.5).
- For the interval (-2.5,∞), we are going to use the test value -2.4
[tex] f'(x)=2x+5 [/tex]
[tex] f'(-2.4)=2(-2.4)+5 [/tex]
[tex] f'(-2.4)=-4.8+5 [/tex]
[tex] f'(-2.4)=0.2 [/tex]
Since the derivative is positive over interval (-2.5,∞), the function [tex] f(x)=x^2+5x+6 [/tex] is increasing over the interval (-2.5,∞).
We can conclude that the graph of f(x) = x2 + 5x + 6 is increasing over the interval (-2.5,∞); therefore none of your given choices is the correct answer.
