consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=\\frac{1}{2}x^{4}-4x^{2}+4$\n\n(a)…

consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=\\frac{1}{2}x^{4}-4x^{2}+4$\n\n(a) find the interval(s) of increase. (enter your answer using interval notation.)\n\nfind the interval(s) of decrease. (enter your answer using interval notation.)\n\n(b) find the local minimum value(s). (enter your answers as a comma - separated list.)\n\nfind the local maximum value(s). (enter your answers as a comma - separated list.)\n\n(c) find the inflection points.\n\nsmaller x - value $(x,y)=$( )\n\nlarger x - value $(x,y)=$( )\n\nfind the interval(s) where the function is concave up. (enter your answer using interval notation.)\n\nfind the interval(s) where the function is concave down. (enter your answer using interval notation.)

consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=\\frac{1}{2}x^{4}-4x^{2}+4$\n\n(a) find the interval(s) of increase. (enter your answer using interval notation.)\n\nfind the interval(s) of decrease. (enter your answer using interval notation.)\n\n(b) find the local minimum value(s). (enter your answers as a comma - separated list.)\n\nfind the local maximum value(s). (enter your answers as a comma - separated list.)\n\n(c) find the inflection points.\n\nsmaller x - value $(x,y)=$( )\n\nlarger x - value $(x,y)=$( )\n\nfind the interval(s) where the function is concave up. (enter your answer using interval notation.)\n\nfind the interval(s) where the function is concave down. (enter your answer using interval notation.)

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ( (x^n)^\prime=nx^{n - 1} ), for ( f(x)=\frac{1}{2}x^{4}-4x^{2}+4 ), we have ( f^\prime(x)=2x^{3}-8x = 2x(x^{2}-4)=2x(x - 2)(x + 2) )

Step2: Determine intervals of increase and decrease

Set ( f^\prime(x)=0 ), then ( x=-2,0,2 ).

  • For ( x\in(-2,0) ), let ( x=-1 ), ( f^\prime(-1)=2\times(-1)\times((-1)^{2}-4)=6>0 ) (wrong, actually ( f^\prime(-1)=2\times(-1)\times(1 - 4)=6>0 ) is wrong, correct: ( f^\prime(-1)=2\times(-1)\times((-1)^{2}-4)=2\times(-1)\times(-3) = 6>0 ) (no, correct: ( f^\prime(x)=2x(x - 2)(x + 2) ), when ( x=-1 ), ( f^\prime(-1)=2\times(-1)\times(-1 - 2)\times(-1 + 2)=2\times(-1)\times(-3)\times1=6>0 ) (no, correct: ( f^\prime(x)=2x^{3}-8x ), ( f^\prime(-1)=2\times(-1)^{3}-8\times(-1)=-2 + 8=6>0 )).
  • For ( x\in(-\infty,-2) ), let ( x=-3 ), ( f^\prime(-3)=2\times(-3)^{3}-8\times(-3)=-54 + 24=-30<0 )
  • For ( x\in(0,2) ), let ( x = 1 ), ( f^\prime(1)=2\times1^{3}-8\times1=2-8=-6<0 )
  • For ( x\in(2,\infty) ), let ( x = 3 ), ( f^\prime(3)=2\times3^{3}-8\times3=54-24 = 30>0 )

Intervals of increase: ( (-2,0)\cup(2,\infty) ) Intervals of decrease: ( (-\infty,-2)\cup(0,2) )

Step3: Find local minima and maxima

Using the first - derivative test:

  • At ( x=-2 ), ( f^\prime(x) ) changes from negative to positive. ( f(-2)=\frac{1}{2}\times(-2)^{4}-4\times(-2)^{2}+4=\frac{1}{2}\times16-4\times4 + 4=8-16 + 4=-4 )
  • At ( x = 0 ), ( f^\prime(x) ) changes from positive to negative. ( f(0)=\frac{1}{2}\times0^{4}-4\times0^{2}+4=4 )
  • At ( x = 2 ), ( f^\prime(x) ) changes from negative to positive. ( f(2)=\frac{1}{2}\times2^{4}-4\times2^{2}+4=\frac{1}{2}\times16-4\times4 + 4=8-16 + 4=-4 )

Local minimum values: ( -4,-4 ) Local maximum value: ( 4 )

Step4: Find the second derivative

( f^\prime(x)=2x^{3}-8x ), then ( f^{\prime\prime}(x)=6x^{2}-8=2(3x^{2}-4) ) Set ( f^{\prime\prime}(x)=0 ), ( 3x^{2}-4 = 0), ( x=\pm\frac{2}{\sqrt{3}}=\pm\frac{2\sqrt{3}}{3})

  • ( f(\frac{2\sqrt{3}}{3})=\frac{1}{2}\times(\frac{2\sqrt{3}}{3})^{4}-4\times(\frac{2\sqrt{3}}{3})^{2}+4=\frac{1}{2}\times\frac{16\times9}{81}-4\times\frac{12}{9}+4=\frac{8}{9}-\frac{16}{3}+4=\frac{8 - 48 + 36}{9}=-\frac{4}{9})
  • ( f(-\frac{2\sqrt{3}}{3})=\frac{1}{2}\times(-\frac{2\sqrt{3}}{3})^{4}-4\times(-\frac{2\sqrt{3}}{3})^{2}+4=-\frac{4}{9})

Inflection points: ( (-\frac{2\sqrt{3}}{3},-\frac{4}{9}),(\frac{2\sqrt{3}}{3},-\frac{4}{9}) )

Step5: Determine concavity

  • For ( x\in(-\infty,-\frac{2\sqrt{3}}{3})\cup(\frac{2\sqrt{3}}{3},\infty) ), let ( x=-2 ), ( f^{\prime\prime}(-2)=6\times(-2)^{2}-8=24 - 8 = 16>0 ); let ( x = 2 ), ( f^{\prime\prime}(2)=6\times2^{2}-8=24 - 8=16>0 ). Concave up: ( (-\infty,-\frac{2\sqrt{3}}{3})\cup(\frac{2\sqrt{3}}{3},\infty) )
  • For ( x\in(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}) ), let ( x = 0 ), ( f^{\prime\prime}(0)=6\times0^{2}-8=-8<0 ). Concave down: ( (-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}) )

Answer:

(a) Intervals of increase: ( (-2,0)\cup(2,\infty) ); Intervals of decrease: ( (-\infty,-2)\cup(0,2) ) (b) Local minimum values: ( -4,-4 ); Local maximum value: ( 4 ) (c) Inflection points: ( (-\frac{2\sqrt{3}}{3},-\frac{4}{9}),(\frac{2\sqrt{3}}{3},-\frac{4}{9}) ); Concave up: ( (-\infty,-\frac{2\sqrt{3}}{3})\cup(\frac{2\sqrt{3}}{3},\infty) ); Concave down: ( (-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}) )