Consider the function below. (If an answer does not exist, enter DNE.) f(x) = 1/
ID: 2883074 • Letter: C
Question
Consider the function below. (If an answer does not exist, enter DNE.) f(x) = 1/2 x^4 - 4x^2 + 2 (a) Find the interval of increase. (Enter your answer using interval notation.) Find the interval of decrease. (Enter your answer using interval notation.) (b) Find the local minimum value(s). (Enter your answers as a comma-separated list.) Find the local maximum value(s). (Enter your answers as a comma-separated list.) (c) Find the inflection points. Find the interval where the graph is concave upward. (Enter your answer using interval notation.) Find the interval where the graph is concave downward. (Enter your answer using interval notation.)Explanation / Answer
given f(x)=(1/2)x4-4x2+2, domain is (-,)
differentiate with respect to x
f '(x)=2x3-8x
differentiate with respect to x
f "(x)=6x2-8
a) f(x) increases when f '(x)>0
2x3-8x>0
2(x+2)x(x-2)>0
(-2,0)U(2,)
f(x) decreases when f '(x)<0
2x3-8x<0
2(x+2)x(x-2)<0
(-,-2)U(0,2)
b)f "(-2)=6(-2)2-8=16>0
f "(0)=6(0)2-8=-8<0
f "(2)=6(2)2-8=16>0
local minimum at x =-2 ,2
local maximum at x=0
f(-2)=-6,f(0)=2,f(2)=-6
local minimum value =-6
local maximum value =2
c)f "(x)=6x2-8
for inflection point f "(x)=0
6x2-8=0
x2=4/3
x=-2/3 ,2/3
f(-2/3) =(8/9)-(16/3)+2=-22/9
f(2/3) =(8/9)-(16/3)+2=-22/9
inflection point at (-2/3,-22/9), (2/3,-22/9)
concave upward when f "(x)>0
6x2-8>0
(-,-2/3)U(2/3,)
concave downward when f "(x)<0
6x2-8<0
(-2/3,2/3)