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

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. $f ( - 6 ) = 0$

Quizplus · Accepted Answer

To find the intervals of increasing and decreasing, we need to look at the first derivative of the function, which is $f'(x) = 2x + 6$. Setting this derivative equal to zero, we get $2x + 6 = 0$, which gives us $x = -3$. So, $f'(x) > 0$ for $x < -3$, meaning $f$ is increasing on $( - \infty , - 3 )$. Similarly, $f'(x) < 0$ for $x > -3$, meaning $f$ is decreasing on $( - 3 , \infty )$. To find the relative extrema, we look at the original function at the critical point $x = -3$ and the endpoints of the intervals. $f(-3) = (-3)^2 + 6(-3) = -9$ is a relative minimum.

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

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