Question

Consider the function f(x)=1x2+6x+15 a) Give the domain of f (in interval notation) equation editorEquation Editor b) Find the critical numbers of f. equation editorEquation Editor (Separate multiple answers by commas.) c) Determine the intervals on which f is increasing and decreasing. f is increasing on: equation editorEquation Editor f is decreasing on: equation editorEquation Editor d) Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. If the function does not have a relative max or min, type none Relative maxima occur at x= equation editorEquation Editor (Separate multiple answers by commas.) Relative minima occur at x= equation editorEquation Editor (Separate multiple answers by commas.) Note: You can earn partial credit on this proble

Answer 1

(a) The domain of the function is
`(-\infty, \infty)`

(b) The critical number of the function is
`x=-3`

(c) The function is decreasing on the interval
`(-\infty, -3)` and it is increasing on the interval
`(-3, \infty)`

(d)
`f(x)` has a relative minimum point at x = -3

Explanation:

We have the following function
`f(x)=1x^2+6x+15` 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:
`(-\infty, \infty)`

(b) We say that
`x=c` is a critical number of the function
`f(x)` if
`f(c)` exists and if either of the following are true.

`f'(c)=0 \quad OR \quad f'(c) \quad {doesn't \:exist}`

We first need the derivative of the function
`f(x)=1x^2+6x+15`

`(d)/(dx) f(x)=(d)/(dx)(1x^2+6x+15)\\\\f'(x)=2x+6`

the only critical points will be those values of x which make the derivative zero. So, we must solve

`2x+6=0\\x=-3`

(c) To determine the intervals of increase and decrease of the function
`f(x)=1x^2+6x+15`, 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
`(-\infty, -3)`, take x = -4

`f'(-4)=2(-4)+6=-2`

f'(x) < 0 therefore f(x) is decreasing

On the interval
`(-3, \infty)`, take x = 0

`f'(0)=2(0)+6=6` f'(x) > 0 therefore f(x) is increasing

The function is decreasing on the interval
`(-\infty, -3)` and it is increasing on the interval
`(-3, \infty)`

(d) An extremum point would be a point where
`f(x)` is defined and
`f' (x)` change signs.

`f(x)` decreases (f'(x) < 0) before x = -3, increases after it (f'(x) > 0). So
`f(x)` has a relative minimum point at x = -3

We can check our work with the graph of the function.