What are the absolute extrema of f(x)=5x^7 - 7x^5 - 5 in[-oo,oo]?

2 Answers
Jan 26, 2016

There are not absolute extrema because f(x) unbounded
There are local extrema:

LOCAL MAX: x=-1
LOCAL MIN: x=1
INFLECTION POINT x=0

Explanation:

There are not absolute extrema because

lim_(x rarr +-oo) f(x) rarr +-oo

You could find local extrema, if any.

To find f(x) extrema or critical poits we have to computate f'(x)

When f'(x)=0 => f(x) has a stationary point (MAX, min or inflection point).
Then we have to find when:

f'(x)>0 => f(x) is increasing
f'(x)<0 => f(x) is decreasing

Therefore:

f'(x)=d/dx(5x^7-7x^5-5)=35x^6-35x^4+0=35x^4(x^2-1)

:.f'(x)=35x^4(x+1)(x-1)

  • f'(x)=0

color(green)cancel(35)x^4(x+1)(x-1)=0

x_1=0
x_(2,3)=+-1

  • f'(x)>0

x^4>0 AAx
x+1>0 => x> -1
x-1>0 => x>1

Drawing the plot, you'll find

enter image source here

f'(x)>0 AAx in (-oo,-1)uu(1,+oo)

f'(x)<0 AAx in(-1,1)

:.f(x) increasing AA x in (-oo,-1)uu(1,+oo)
:.f(x) decreasing AA x in (-1,1)

x=-1=>LOCAL MAX
x=+1=> LOCAL MIN
x=0=> INFLECTION POINT

graph{5x^7-7x^5-5 [-16.48, 19.57, -14.02, 4]}

Jan 26, 2016

That function has no absolute extrema.

Explanation:

lim_(xrarroo)f(x) = oo and lim_(xrarr-oo)f(x) = -oo.

So the function is unbounded in both directions.