Consider a function f(x) such that f(x+1) = x^2 + 4x + k and f(2) = 8.
a) Find the value of k.
b) Find f(-1)
c) Express f(x) in terms of x
d) Hence solve f(x) = 3
maths...
2006-10-15 2:50 am
回答 (3)
2006-10-15 2:56 am
✔ 最佳答案
(a) f(x + 1) = x^2 + 4x + kf(1 + 1) = 1^2 + 4(1) + k
f(2) = 1 + 4 + k
8 = 5 + k
k = 3.
(b) f(-1) = f(-2 + 1)
= (-2)^2 + 4(-2) + k
= 4 - 8 + 3
= -1
(c) Put y = x + 1. Then x = y - 1
Then f(y)
= f(x + 1) = x^2 + 4x + 3
= (y - 1)^2 + 4(y - 1) + 3
= y^2 - 2y + 1 + 4y - 4 + 3
= y^2 + 2y.
Thus f(x) = x^2 + 2x
(d) f(x) = 3
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or 1.
2006-10-15 4:00 am
(a) f(x + 1) = x^2 + 4x + k
f(1 + 1) = 1^2 + 4(1) + k
f(2) = 1 + 4 + k
8 = 5 + k
k = 3.
(b) f(-1) = f(-2 + 1)
= (-2)^2 + 4(-2) + k
= 4 - 8 + 3
= -1
(c) Put y = x + 1. Then x = y - 1
Then f(y)
= f(x + 1) = x^2 + 4x + 3
= (y - 1)^2 + 4(y - 1) + 3
= y^2 - 2y + 1 + 4y - 4 + 3
= y^2 + 2y.
Thus f(x) = x^2 + 2x
(d) f(x) = 3
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or 1.
f(1 + 1) = 1^2 + 4(1) + k
f(2) = 1 + 4 + k
8 = 5 + k
k = 3.
(b) f(-1) = f(-2 + 1)
= (-2)^2 + 4(-2) + k
= 4 - 8 + 3
= -1
(c) Put y = x + 1. Then x = y - 1
Then f(y)
= f(x + 1) = x^2 + 4x + 3
= (y - 1)^2 + 4(y - 1) + 3
= y^2 - 2y + 1 + 4y - 4 + 3
= y^2 + 2y.
Thus f(x) = x^2 + 2x
(d) f(x) = 3
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or 1.
2006-10-15 3:02 am
a. if f(2) = 8 then x = 1
8 = 1+ 4 + k
k = 3
b. if f(-1) then x = -2
f(-1) = (-2)^2 + 4(-2) + 3
= 4 - 8 + 3
= -1
c. f(x) => x = x +1
f(x) = (x+1)^2 + 4(x+1) + 3
= x^2 + 2x + 1 + 4x + 4 + 3
= x^2 + 6x + 8
d. x^2 + 6x + 8 = 3
x^2 + 6x + 5 = 0
(x + 1)(x + 5) = 0
x = -1 or x = -5
8 = 1+ 4 + k
k = 3
b. if f(-1) then x = -2
f(-1) = (-2)^2 + 4(-2) + 3
= 4 - 8 + 3
= -1
c. f(x) => x = x +1
f(x) = (x+1)^2 + 4(x+1) + 3
= x^2 + 2x + 1 + 4x + 4 + 3
= x^2 + 6x + 8
d. x^2 + 6x + 8 = 3
x^2 + 6x + 5 = 0
(x + 1)(x + 5) = 0
x = -1 or x = -5
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