Jaká je rovnice tečné přímky f (x) = e ^ (x) / (x ^ 2-x při x = 3?

Odpovědět:

# y = e ^ 3 / 36x + e ^ 3/12 #

#f (x) = e ^ x / (x ^ 2-x) #

# D_f = {AAx ##v## RR ##: x ^ 2-x! = 0} = (- oo, 0) uu (0,1) uu (1, + oo) = RR- {0,1} #

#f '(x) = (e ^ x / (x ^ 2-x))' = ((e ^ x) '(x ^ 2-x) -e ^ x (x ^ 2-x)') / (x ^ 2-x) ^ 2 = #

# (e ^ x (x ^ 2-x) -e ^ x (2x-1)) / (x ^ 2-x) ^ 2 = (x ^ 2e ^ x-xe ^ x-2xe ^ x + e ^ x) / (x ^ 2-x) ^ 2 = #

# (x ^ 2e ^ x-3xe ^ x + e ^ x) / (x ^ 2-x) ^ 2 #

Pro rovnici tečny na #A (3, f (3)) # požadujeme hodnoty

#f (3) = e ^ 3/6 #

#f '(3) = (9e ^ 3-9e ^ 3 + e ^ 3) / 36 = e ^ 3/36 #

Rovnice bude

# y-f (3) = f '(3) (x-3) # #<=>#

# y-e ^ 3/6 = e ^ 3/36 (x-3) # #<=>#

# y-e ^ 3/6 = e ^ 3 / 36x-zrušit (3) e ^ 3 / zrušit (36) # #<=>#

# y = e ^ 3 / 36x-e ^ 3/12 + e ^ 3/6 # #<=>#

# y = e ^ 3 / 36x + e ^ 3/12 #

a graf