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在2012届广东省惠州市第三次模拟考试中,有一道考察“导数”的试题,内容如下:已知函数f(x)=x3-3x.(1)求曲线y=f(x)在点x=2处的切线方程;(2)若过点A(1,m)(m≠-2)可作曲线y=f(x)的三条切线,求实数m的取值范围.解:(1)略;(2)过点A(1,m)向曲线y=f(x)作切线,设切点为(x0,y0)则y0=x30-3x0,k′(x0)=3x20-3.则切线方程为y-(x30-3x0)=(3x20-3)(x-x0)将A(1,m)代入上式,整理得2x30-3x20+m+
In 2012 Third Examination of Huizhou City, Guangdong Province, there is an examination of the “Derivatives” test questions, as follows: Known function f (x) = x3-3x. (1) Find the curve y = f ) At the point x = 2 tangent equation; (2) If the point A (1, m) (m ≠ -2) can be used for the three tangents of the curve y = f (x), find the range of real numbers m. Solution: (1) Slightly; (2) If point A (1, m) is tangential to the curve y = f (x) and the point of contact is (x0, y0) then y0 = x30-3x0, k ’ = 3x20-3. Then the tangent equation is substituted by y (x30-3x0) = (3x20-3) (x-x0) into the above formula, and the result is 2x30-3x20 + m +