consider the following. (if an answer does not exist, enter dne.)\n$f(x)=x^{3}+3 x^{2}-72 x$\n(a) find the…

consider the following. (if an answer does not exist, enter dne.)\n$f(x)=x^{3}+3 x^{2}-72 x$\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum value of $f$.\nlocal minimum value\n\nlocal maximum value\n

consider the following. (if an answer does not exist, enter dne.)\n$f(x)=x^{3}+3 x^{2}-72 x$\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum value of $f$.\nlocal minimum value\n\nlocal maximum value\n

Answer

Explanation:

Step1: Find the derivative of (f(x))

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (f(x)=x^{3}+3x^{2}-72x), we have (f^\prime(x)=3x^{2}+6x - 72). Factor out a (3): (f^\prime(x)=3(x^{2}+2x - 24)). Then factor the quadratic: (f^\prime(x)=3(x + 6)(x - 4)).

Step2: Find the critical points

Set (f^\prime(x)=0). So (3(x + 6)(x - 4)=0). Using the zero - product property (x+6 = 0) or (x - 4=0). The critical points are (x=-6) and (x = 4).

Step3: Determine the sign of (f^\prime(x)) in the intervals

We consider the intervals ((-\infty,-6)), ((-6,4)), and ((4,\infty)).

  • For (x\in(-\infty,-6)), let (x=-7). Then (f^\prime(-7)=3(-7 + 6)(-7 - 4)=3(-1)(-11)=33>0).
  • For (x\in(-6,4)), let (x = 0). Then (f^\prime(0)=3(0 + 6)(0 - 4)=3\times6\times(-4)=-72<0).
  • For (x\in(4,\infty)), let (x = 5). Then (f^\prime(5)=3(5 + 6)(5 - 4)=3\times11\times1 = 33>0).

Step4: Find the intervals of increase and decrease

Since (f^\prime(x)>0) on ((-\infty,-6)\cup(4,\infty)), the function (f(x)) is increasing on ((-\infty,-6)\cup(4,\infty)). Since (f^\prime(x)<0) on ((-6,4)), the function (f(x)) is decreasing on ((-6,4)).

Step5: Find the local minimum and maximum values

We use the second - derivative test. The second derivative (f^{\prime\prime}(x)=6x + 6).

  • For (x=-6), (f^{\prime\prime}(-6)=6\times(-6)+6=-30<0). So (f(-6)=(-6)^{3}+3(-6)^{2}-72(-6)=-216 + 108+432=324) is a local maximum.
  • For (x = 4), (f^{\prime\prime}(4)=6\times4+6=30>0). So (f(4)=4^{3}+3\times4^{2}-72\times4=64 + 48-288=-176) is a local minimum.

Answer:

(a) ((-\infty,-6)\cup(4,\infty)) (b) ((-6,4)) (c) local minimum value: (-176), local maximum value: (324)