Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4
ID: 2891500 • Letter: C
Question
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 - 50x2 + 2 (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval 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 points. (x, y) = ( (smaller x-value) (x, y) = ( (larger x-value) Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down.(Enter your answer using interval notation.)Explanation / Answer
We have given f(x)=x^4-50x^2+2
f'(x)=4x^3-100x=0
x(4x^2-100)=0
x=0,4x^2-100=0 implies x^2=25 implies x=5,-5
we have critical points x=0,5,-5
f'(-6)=4(-6)^3-100*(-6)=-264<0
f'(-1)=4(-1)^3-100*(-1)=96>0
f'(1)=4(1)^3-100*(1)=-96<0
f'(-4)=4(-4)^3-100*(-4)=144>0
f'(6)=4(6)^3-100*(6)=264>0
a) f is increasing on the interval (-5,0) u (5,+infinity)
f is decreasing on the interval (-infinity,-5) u (0,5)
b) f'(-1)=4(-1)^3-100*(-1)=96>0 left of the x=0 and f'(1)=4(1)^3-100*(1)=-96<0 to the right of x=0
then x=0 is a local maximum is f(0)=0-0+2=2
f'(4)=4(4)^3-100*(4)=-144<0 left of the x=5 and f'(6)=4(6)^3-100*(6)=264>0 to the right of x=5
then x=5 is a local minimum is f(5)=(5)^4-50*(5)^2+2=-623
f'(-6)=4(-6)^3-100*(-6)=-264<0 left of the x=-5 and f'(-4)=4(-4)^3-100*(-4)=144>0 to the right of x=-5
then x=-5 is a local minimum is f(-5)=(-5)^4-50*(-5)^2+2=-623
local minimum values is at x=5,-5 is (5,-623) and (-5,-623)
local maximum values is at x=0 is 2
c) we have f'(x)=4x^3-100x
f''(x)=12x^2-100
set f''(x)=0
12x^2-100=0
12x^2=100
x^2=100/12=50/6=25/3
x=5/sqrt(3),x=-5/sqrt(3)
f''(x)=12x^2-100
f''(3)=12(3)^2-100=8>0
f''(2)=12(2)^2-100=-52<0
f''(x) changes signs at x=5/sqrt(3) negative to positive
f''(-3)=12(-3)^2-100=8>0
f''(-2)=12(-2)^2-100=-52<0
f''(x) changes signs at x=-5/sqrt(3) positive to negative
so inflection points are x=(-5/sqrt(3),-345.222222222),(5/sqrt(3),-345.222222222)
for smaller x-value (x,y)=(-5/sqrt(3),-345.222222222)
for larger x-value (x,y)=(5/sqrt(3),-345.222222222)
f''(x) >0 on the interval (-infinity,-5/sqrt(3)) and (5/sqrt(3),+infinity)
so f is concave up on the interval (-infinity,-5/sqrt(3)) and (5/sqrt(3),+infinity)
f''(x) <0 on the interval (-5/sqrt(3),5/sqrt(3))
so f is concave down on the interval (-5/sqrt(3),5/sqrt(3))