given the function g(x)= x^3-2x+3, the differnce quotiebt (g(1+h)-g(1)/h. (h dont equl 0) is equal to?
0 or h+1 or h?
given the function g(x)= x^3-2x+3, the differnce quotiebt (g(1+h)-g(1)/h. (h dont equl 0) is equal to?
2018-03-12 2:44 pm
回答 (2)
2018-03-12 3:51 pm
g(x) = x³ - 2x + 3
g(x + h)
= (x + h)³ - 2(x + h) + 3
= x³ + 3x²h + 3xh² + h³ - 2x - 2h + 3
[g(1 + h) - g(1)] / h
= {[(1)³ + 3h(1)² + 3h²(1) + h³ - 2(1) - 2h + 3] - [(1)³ - 2(1) + 3]} / h
= (1 + 3h + 3h² + h³ - 2 - 2h + 3 - 1 + 2 - 3) / h
= (h³ + 3h² + h) / h
= h² + 3h + 1
g(x + h)
= (x + h)³ - 2(x + h) + 3
= x³ + 3x²h + 3xh² + h³ - 2x - 2h + 3
[g(1 + h) - g(1)] / h
= {[(1)³ + 3h(1)² + 3h²(1) + h³ - 2(1) - 2h + 3] - [(1)³ - 2(1) + 3]} / h
= (1 + 3h + 3h² + h³ - 2 - 2h + 3 - 1 + 2 - 3) / h
= (h³ + 3h² + h) / h
= h² + 3h + 1
2018-03-12 7:56 pm
g(x) =x^3-2x+3
g(1) = 1^3-2(1)+3 = 2
g(1+h) = (1+h)^3-2(1+h)+3
g(1+h) = 1+3h+3h^2+h^3 -2-2h+3
g(1) = 2
g(1+h)-g(1) = ( 1+3h+3h^2+h^3 -2-2h+3 -2)
g(1+h)-g(1) = ( 3h+3h^2+h^3-2h)
(g(1+h)-g(1))/h = 3+3h+h^2-2
(g(1+h)-g(1))/h = 3h+h^2+1
g(1) = 1^3-2(1)+3 = 2
g(1+h) = (1+h)^3-2(1+h)+3
g(1+h) = 1+3h+3h^2+h^3 -2-2h+3
g(1) = 2
g(1+h)-g(1) = ( 1+3h+3h^2+h^3 -2-2h+3 -2)
g(1+h)-g(1) = ( 3h+3h^2+h^3-2h)
(g(1+h)-g(1))/h = 3+3h+h^2-2
(g(1+h)-g(1))/h = 3h+h^2+1
收錄日期: 2021-04-24 01:08:32
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https://hk.answers.yahoo.com/question/index?qid=20180312064428AAxlCqr