F (x) = x & # 179; - 3x & # 178; + 3ax-3a + 3 (1) find the tangent equation of the curve y = f (x) at (1, f (1)) (2) x belongs to [0,2], Find the maximum value of | f (x) |

F (x) = x & # 179; - 3x & # 178; + 3ax-3a + 3 (1) find the tangent equation of the curve y = f (x) at (1, f (1)) (2) x belongs to [0,2], Find the maximum value of | f (x) |

（1）
f '(x)=3x²-6x+3a
Slope k = F & nbsp;, # 39; (1) = 3 (A-1)
Ordinate of tangent point: F (1) = 1
Tangent equation:
y-1=3(a-1)(x-1)
(2)
f '(x)=3(x²-2x+a)
If Δ x ≤ 0 = = & gt; 4 ≤ 4a, i.e. a ≥ 1, the function increases monotonically on [0,2]
|f(0)|=|3-3a|=3a-3
|f(2)|=|3a-1|=3a-1
(|fx|)max=3a-1
Let F & nbsp; ' (x) = 0 = = & gt; X (12) = 1 ± √ (1-A)
(x-1)²=1-a
X & amp; # 178; = 2x-a (exchange conditions between X & amp; # 178; and x)
X & amp; # 179; - 3x & amp; # 178; = x & amp; # 178; (x-3) = (2x-a) (x-3) = 2x & amp; # 178; - (a + 3) x + 3A = 2 (2x-a) - (a + 3) x + 3A = x-ax + a (f (x)) cold treatment of the first two items)
f(x)=x-ax+a+3ax-3a+3=(2a+1)x+(3-2a)
Note: the f (x) in the previous line is not the f (x) of the original function, but the f (x) of the special extremum
The G (x) in the picture is the f (x) here. If we don't, we can't distinguish it,
The purpose of introducing g (x) is to reduce the amount of computation, otherwise the extreme point is very difficult to calculate;

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