Hanapin ang saklaw ng function f (x) = (1+ x ^ 2) / x ^ 2?

Sagot:

#f (A) = (1, oo) #

Paliwanag:

#f (x) = (x ^ 2 + 1) / x ^ 2 #, #A = (- oo, 0) uu (0, oo) #

#f '(x) = ((x ^ 2 + 1)' x ^ 2 (x ^ 2) '(x ^ 2 + 1)) / x ^ 4 =

# (2x ^ 3-2x ^ 3-2x) / x ^ 4 = #

# -2 / x ^ 3 #

Para sa #x> 0 # meron kami #f '(x) <0 # kaya nga # f # ay mahigpit na bumababa sa # (0, oo) #

Para sa #x <0 # meron kami #f '(x)> 0 # kaya nga # f # ay mahigpit na tumataas # (- oo, 0) #

# A_1 = (- oo, 0) #, # A_2 = (0, oo) #

(x) = lim (xrarr0 ^ (-

#lim_ (xrarr0 ^ (+)) f (x) = lim_ (xrarr0 ^ (+)) (x ^ 2 + 1) / x ^ 2 =

# xl (xrarr-oo) f (x) = lim_ (xrarr-oo) (x ^ 2 + 1) / x ^ 2 = lim_ (xrarr-oo) x ^ 2 /

#lim_ (xrarr + oo) f (x) = lim_ (xrarr + oo) (x ^ 2 + 1) / x ^ 2 = 1 #

#f (x) = f (((- oo, 0))) = (lim_ (xrarr-oo) f (x), lim_ (xrarr0 ^ (-)) f (x)

# (1, oo) #

# x (A_2) = f (((0, + oo))) = (lim_ (xrarr + oo) f (x), lim_ (xrarr0 ^ +) f (x)) = (1,

Saklaw # = f (A) = f (A_1) uuf (A_2) = (1, oo) #