Cari julat fungsi f (x) = (1+ x ^ 2) / x ^ 2?

Jawapan:

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

Penjelasan:

#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 #

Untuk #x> 0 # kita ada #f '(x) <0 # jadi # f # sangat ketat dalam # (0, + oo) #

Untuk #x <0 # kita ada #f '(x)> 0 # jadi # f # semakin ketat # (- oo, 0) #

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

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

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

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

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

f (x), f (x), f (x)

# (1, + oo) #

f (x) = (1, + oo) #f (A_2) = f (((0,

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