Carefully skecth the graph of f(x)= x^4-6x^2, clearly showing and labeling the x & y coordinates of all absolute extrema, all local extrema... Carefully skecth the graph of f(x)= x^4-6x^2, clearly showing and labeling the x & y coordinates of all absolute extrema, all local extrema, and all inflection points . Justify your answer using calculus.
- print Print
- list Cite
Expert Answers
briefcaseTeacher (K-12)
calendarEducator since 2011
write3,149 answers
starTop subjects are Math, Science, and Business
Given `f(x)=x^4-6x^2` :
`f'(x)=4x^3-12x`
`f''(x)=12x^2-12`
(a) All extrema occur at critical points (i.e. where f'(x)=0 or f'(x) fails to exist.) Since this is a polynomial, we are only concerned with finding points where f'(x)=0:
`4x^3-12x=0==>4x(x^2-3)=0` so x=0 or `x=+-sqrt(3)`
The graph of a quartic with positive leading coefficient has end behavior of `lim_(x->+-oo)=oo` so there will be no absolute maximum. There will be an absolute minimum.
We test values of f'(x) on the following intervals:
`(-oo,-sqrt(3):` f'(x)<0 so the function is decreasing on this interval.
`f(-sqrt(3))=-9`
`(-sqrt(3),0):` f'(x)>0 so the function is increasing on this interval.
f(0)=0
`(0,sqrt(3)):` f'(x)<0 so the function is decreasing on this interval
`f(sqrt(3))=-9`
`(sqrt(3),oo):` f'(x)>0 so the function is increasing on this interval.
There is a local maximum at x=0 with value 0, and the absolute minimum for the function occurs at `x=-sqrt(3),x=sqrt(3)` with value -9.
(b) Points of inflection occur when the concavity of the graph changes at a point -- i.e. the sign of the secon derivative changes at the point.
`12x^2-12=0==>x=+-1` . We check the sign of the second derivative on the following intervals:
`(-oo,-1):` f''(x)>0 so the function is concave up on this interval.
`(-1,1):` f''(x)<0 so the function is concave down on this interval
`(1,oo):` f''(x)>0 so the function is concave up on this interval.
The points of inflection are at x=-1 and x=1.
The y-intercept is at 0; the x-intercepts are `x=0,+-sqrt(6)`
The graph:
Related Questions
- `f(x) = 2x^3 - 6x, [0,3]` Find the absolute extrema of the function on the closed interval.
- 2 Educator Answers
- `f(x) = x^3 - (3/2)x^2, [-1,2]` Find the absolute extrema of the function on the closed...
- 1 Educator Answer
- Find the absolute maximum and absolute minimum values of the function f(x)= (e^(-6x))-(e^(-5x))on...
- 1 Educator Answer
- What is the difference between local max and absolute max ? sketch the graph. a. the domain...
- 1 Educator Answer
- `f(x) = (2x)/(x^2 + 1), [-2,2]` Find the absolute extrema of the function on the closed...
- 2 Educator Answers