8.y' = 2x
f'(2) = 4
所以切梁弊线斜率为4
用点斜式:y = 4(x - 2) + 4
y = 4x - 8 + 4
y = 4x - 4
所以切线方程 y = 4x - 4
9.f'(x) = 3x^2 - 6x - 9
= 3(x^2 - 2x - 3)
= 3(x - 3)(x +1)
x (-∞,-1) -1 (-1,3) 3 (3,+∞)
y' + 0 - 0 +
y ↗ ↘ ↗
所以单调递减区间为(缺渣悔-1,3)
10.因为 g(1) = 3
所以伏正 f[g(1)] = f(3) = 1
因为 g[f(x)] = 2
所以 f(x) = 2
x = 1
1.求导激兄,可算出斜率为4,所以切线举察方程:Y=4X-4
2.求导,3x^2-6x-9<0区间(明答袭-1,3)
3.1;x=1
1.y=2*2(x-2)+4=4x-4
2.导陪族州数=3x²-6x-9<0 =====>芦蔽穗指-1
g(2)=2,f(1)=2,即x=1