Critical Numbers and Relative Extrema

[lovetolearn]

Let f''(x)=4x³-2x and let f(x) have the critical numbers -1, 0, and
1. Determine if any of the critical numbers gives a relative extrema.
Justify your answer(s).






f'(x) = x^4 - x^2 + C
f''(0) = 0 --> inflection
f''(1) = 2, f''(-1) = -6, not inflections
f'(-1) and f'(1) = 0 --> local extrema
if they're not inflections, they're min or max
if slope = 0
also, C = 0, given the critical numbers
max at -1, min at +1
f(x) =(x^5/5)- (x^3/3) + C




OR IS IT




f'(x)=x^4-x^2
f(x)=((5x^2)/(5))-((x^3)/(3))+3x+alpha
at critical points c=0
inflection at 0
f(-1)=(-1/5)+(1/3)+d
f(0)=d
f(1)=(1/5)-(1/3)+d
max at -1 min at +1
max at (-1, 2/15) if c=0
min at (1, -2/15)


OR am I completely wrong?

 

[HallsofIvy]

Let f''(x)=4x³-2x and let f(x) have the critical numbers -1, 0, and
1. Determine if any of the critical numbers gives a relative extrema.
Justify your answer(s).






f'(x) = x^4 - x^2 + C

There is no point in calculating this. You were told that f has "critical numbers -1, 0, and 1" so of course the derivative must be 0.

f''(0) = 0 --> inflection
f''(1) = 2, f''(-1) = -6, not inflections
f'(-1) and f'(1) = 0 --> local extrema
if they're not inflections, they're min or max
if slope = 0
also, C = 0, given the critical numbers
max at -1, min at +1

f(x) =(x^5/5)- (x^3/3) + C

Again, it was not necessary to calclulate this. At any critical point where the derivative is 0, the second derivative being 0 gives an inflection point, positive, a local minimum, and negative, a local maximum.

OR IS IT




f'(x)=x^4-x^2
f(x)=((5x^2)/(5))-((x^3)/(3))+3x+alpha
at critical points c=0
inflection at 0
f(-1)=(-1/5)+(1/3)+d
f(0)=d
f(1)=(1/5)-(1/3)+d
max at -1 min at +1
max at (-1, 2/15) if c=0
min at (1, -2/15)


OR am I completely wrong?

Not wrong but mostly unnecessary. All you really need is that f''(0)= 0, f''(1)> 0, f''(-1)< 0.