Question

consider the equation below. (if an answer does not exist, enter dne.)

$f(x)=x^{3}-12x^{2}-27x + 3$

(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)

find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)

(b) find the local minimum and maximum values of $f$.

local minimum value

local maximum value

(c) find the inflection point.

$(x,y)=$

find the interval(s) on which $f$ is concave up. (enter your answer using interval notation.)

find the interval(s) on which $f$ is concave down. (enter your answer using interval notation.)

Explanation:

Step1: Find the first derivative

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), for \( f(x)=x^{3}-12x^{2}-27x + 3 \), we have \( f^\prime(x)=3x^{2}-24x-27=3(x^{2}-8x - 9)=3(x + 1)(x - 9) \)

Step2: Determine critical points

Set \( f^\prime(x)=0 \), then \( 3(x + 1)(x - 9)=0 \). So \( x=-1 \) or \( x = 9 \)

Step3: Test intervals for increasing/decreasing

• For \( x<-1 \), let \( x=-2 \), \( f^\prime(-2)=3((-2)+1)((-2)-9)=3\times(-1)\times(-11)=33>0 \)
• For \( -1
• For \( x>9 \), let \( x = 10 \), \( f^\prime(10)=3(10 + 1)(10 - 9)=33>0 \)

So \( f(x) \) is increasing on \( (-\infty,-1)\cup(9,\infty) \) and decreasing on \( (-1,9) \)

Step4: Find local extrema

Using the first - derivative test:

• At \( x=-1 \), since \( f(x) \) changes from increasing to decreasing, \( f(-1)=(-1)^{3}-12(-1)^{2}-27(-1)+3=-1-12 + 27+3=17 \) (local maximum)
• At \( x = 9 \), since \( f(x) \) changes from decreasing to increasing, \( f(9)=9^{3}-12\times9^{2}-27\times9+3=729-972-243 + 3=-483 \) (local minimum)

Step5: Find the second derivative

\( f^\prime(x)=3x^{2}-24x-27 \), then \( f^{\prime\prime}(x)=6x-24=6(x - 4) \)

Step6: Determine inflection point and concavity

Set \( f^{\prime\prime}(x)=0 \), then \( 6(x - 4)=0\Rightarrow x = 4 \)

• When \( x = 4 \), \( y=f(4)=4^{3}-12\times4^{2}-27\times4+3=64-192-108 + 3=-233 \)
• For \( x<4 \), let \( x = 0 \), \( f^{\prime\prime}(0)=6(0 - 4)=-24<0 \), so \( f(x) \) is concave down on \( (-\infty,4) \)
• For \( x>4 \), let \( x = 5 \), \( f^{\prime\prime}(5)=6(5 - 4)=6>0 \), so \( f(x) \) is concave up on \( (4,\infty) \)

Answer:

(a) Increasing: \( (-\infty,-1)\cup(9,\infty) \); Decreasing: \( (-1,9) \)
(b) Local minimum value: \( -483 \); Local maximum value: \( 17 \)
(c) Inflection point: \( (4,-233) \); Concave up: \( (4,\infty) \); Concave down: \( (-\infty,4) \)