Answer:
(a) The domain of the function is [tex](-\infty, \infty)[/tex]
(b) The critical number of the function is [tex]x=-3[/tex]
(c) The function is decreasing on the interval [tex](-\infty, -3)[/tex] and it is increasing on the interval [tex](-3, \infty)[/tex]
(d) [tex]f(x)[/tex] has a relative minimum point at x = -3
Step-by-step explanation:
We have the following function [tex]f(x)=1x^2+6x+15[/tex] and we want to find:
(a) The domain of the function is the complete set of possible values of the independent variable.
For this function any real number can be substituted for x and get a meaningful output. Therefore
Domain: [tex](-\infty, \infty)[/tex]
(b) We say that [tex]x=c[/tex] is a critical number of the function [tex]f(x)[/tex] if [tex]f(c)[/tex] exists and if either of the following are true.
[tex]f'(c)=0 \quad OR \quad f'(c) \quad {doesn't \:exist}[/tex]
We first need the derivative of the function [tex]f(x)=1x^2+6x+15[/tex]
[tex]\frac{d}{dx} f(x)=\frac{d}{dx}(1x^2+6x+15)\\\\f'(x)=2x+6[/tex]
the only critical points will be those values of x which make the derivative zero. So, we must solve
[tex]2x+6=0\\x=-3[/tex]
(c) To determine the intervals of increase and decrease of the function [tex]f(x)=1x^2+6x+15[/tex], perform the following:
Form open intervals with critical number and take a value from every interval and find the sign they have in the derivative.
If f'(x) > 0, f(x) is increasing.
If f'(x) < 0, f(x) is decreasing.
On the interval [tex](-\infty, -3)[/tex], take x = -4
[tex]f'(-4)=2(-4)+6=-2[/tex]
f'(x) < 0 therefore f(x) is decreasing
On the interval [tex](-3, \infty)[/tex], take x = 0
[tex]f'(0)=2(0)+6=6[/tex] f'(x) > 0 therefore f(x) is increasing
The function is decreasing on the interval [tex](-\infty, -3)[/tex] and it is increasing on the interval [tex](-3, \infty)[/tex]
(d) An extremum point would be a point where [tex]f(x)[/tex] is defined and [tex]f' (x)[/tex] change signs.
[tex]f(x)[/tex] decreases (f'(x) < 0) before x = -3, increases after it (f'(x) > 0). So [tex]f(x)[/tex] has a relative minimum point at x = -3
We can check our work with the graph of the function.
