For the function is defined by On which two intervals is the function increasing
I got -11 to 6 as one interval
and only part of the other one which is (some # to 10)
What is the "some #" please help this is due tonight. thank you
For the function is defined by On which two intervals is the function increasing
I got -11 to 6 as one interval
and only part of the other one which is (some # to 10)
What is the "some #" please help this is due tonight. thank you
cleaned up your derivative a little bit ...
$\displaystyle f'(x) = 7x^6(x+6)^4 + 4x^7(x+6)^3$
now, look at the following expression ... what can you factor out of those two terms (in other words, what factors do they have in common to "pull" out?)
$\displaystyle 7x^6y^4 + 4x^7y^3$
Hello, Lambo!
The function is increasing when $\displaystyle f'(x)$ is positive.For $\displaystyle x \in [-11, 10]$, the function $\displaystyle f(x)$ is defined by: .$\displaystyle f(x) \:=\:x^7(x+6)^4$
On which two intervals is the function increasing ?
I got -11 to 6 as one interval . . . . sorry, no
The derivative is: .$\displaystyle f'(x) \;=\;x^7\cdot4(x+6)^3 + 7x^6\cdot(x+6)^4 \;=\;x^6(x+6)^3\bigg[4x + 7(x+6)\bigg] $
So we have: .$\displaystyle f'(x)\;=\;x^6\cdot(x+6)^3\cdot(11x+42)\;>\;0$
This product is positive when: .$\displaystyle \begin{Bmatrix}(1) & \text{all three factors are positive} \\ (2) & \text{two factors are negative} \end{Bmatrix}$
(1) All three factors are positive.
. . $\displaystyle x^6$ is always positive, except when $\displaystyle x = 0$
. . $\displaystyle (x+6)^3 \:>\:0\quad\Rightarrow\quad x \:>\:\text{-}6$
. . $\displaystyle 11x+42 \:>\:0 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac{42}{11}$
All three are satisfied if: .$\displaystyle x \:>\:\text{-}\frac{42}{11}\:\text{ and }\:x \neq 0$
. . The interval is: .$\displaystyle \boxed{\;\left(\text{-}\tfrac{42}{11},\:0\right) \cup (0,\:\infty)\;} $
(2) Two factors are negative.
. . $\displaystyle x^6$ is always positive, except when $\displaystyle x = 0$
. . $\displaystyle (x+6)^3 \:<\:0 \quad\Rightarrow\quad x \:<\:\text{-}6$
. . $\displaystyle 11x+42 \:<\:0 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{42}{11}$
All three are satisfied if: .$\displaystyle x \:<\:\text{-}6$
. . The interval is: .$\displaystyle \boxed{\;(\text{-}\infty,\:\text{-}6)\;}$