Given $f ( x ) = x ^ { 2 } + 6 x$.
(a) Find the intervals on which $f$ is increasing or decreasing.
(b) Find the relative maxima and relative minima of $f$.
A) (a) Increasing on $( - 3 , \infty )$,
decreasing on $( - \infty , - 3 )$
(b) Rel. min. $f ( - 3 ) = - 9$
B) (a) Increasing on $( - \infty , - 3 )$,
decreasing on $( - 3 , \infty )$
(b) Rel. max. $f ( - 3 ) = - 9$
C) (a) Increasing on $( - 6 , \infty )$,
decreasing on $( - \infty , - 6 )$
(b) Rel. min. $f ( - 6 ) = 0$
D) (a) Increasing on $( - \infty , - 6 )$,
decreasing on $( - 6 , \infty )$
(b) Rel. max. $f ( - 6 ) = 0$

Question

Quizplus · Accepted Answer

We can find the intervals where the function is increasing or decreasing by taking the derivative of the function and finding its sign. 
$f'(x) = 2x+6$ 
When $f'(x) > 0$, the function is increasing and when $f'(x) < 0$, the function is decreasing. 
$f'(x) = 0$ at $x=-3$. 
Thus, the function is increasing on $( - 3 , \infty )$ and decreasing on $( - \infty , - 3 )$. 
To find the relative maxima and minima, we can look at the concavity of the function. If the second derivative is positive, the function is concave up and there is a relative minimum. If the second derivative is negative, the function is concave down and there is a relative maximum. If the second derivative is zero, we need to use another test.

$f''(x) = 2$ which is always greater than zero. Therefore, the function is concave up and has a relative minimum at $x=-3$. 
Plugging this value into the original function, we get $f(-3) = (-3)^2 + 6(-3) = -9$ as our relative minimum.

Given f(x)=x2+6xf ( x ) = x ^ { 2 } + 6 xf(x)=x2+6x

Given $f ( x ) = x ^ { 2 } + 6 x$