2007-07-17 10:58:46 · 3 answers · asked by wltz4vns 1 in Science & Mathematics ➔ Mathematics
f'(x) = lim (h->0) of [f(x+h) -f(x)] / h
f(x+h) = 3(x+h) + 2 = 3x + 3h + 2
f(x) = 3x + 2
f(x+h) - f(x) = 3x + 3h + 2 - (3x+2)
= 3x + 3h + 2 - 3x - 2
= 3h
[f(x+h) -f(x)] / h
= 3h/h
= 3
f'(x) = lim (h->0) of [f(x+h) -f(x)] / h
= lim (x->0) of 3
= 3
2007-07-17 11:02:10 · answer #1 · answered by MsMath 7 · 2⤊ 0⤋
f `(x) = lim h->0 [ f(x + h) - f(x) ] / h , h â 0
f ` (x) = lim h->0 [ 3(x + h) + 2 - (3x + 2) ] / h
f `(x) = lim h-> 0 [ 3x + 3h + 2 - 3x - 2 ] / h
f `(x) = lim h->0 [ 3h ] / h
f `(x) = 3
2007-07-17 11:12:58 · answer #2 · answered by Como 7 · 1⤊ 0⤋
if u mean by first principle,then:
limh->0 {3(x+h)+2-(3x+2)}/h
Now solv the limit.Quite simple
2007-07-17 11:01:34 · answer #3 · answered by aviral17 3 · 1⤊ 0⤋