8.已知曲线y=f(x)=2x2+4x在点P处的切线斜率为16,则点P的坐标为 .
解析:设P(x0,2x_0^2+4x0),
则f'(x0)=lim┬(Δx"→" 0) (f"(" x_0+Δx")-" f"(" x_0 ")" )/Δx=(lim)┬(Δx"→" 0) (2"(" Δx")" ^2+4x_0 Δx+4Δx)/Δx=4x0+4.
∵f'(x0)=16,∴4x0+4=16,∴x0=3,∴P(3,30).
答案:(3,30)
9.设P为曲线C:f(x)=x2+2x+3上的点,且曲线C在点P处的切线的倾斜角的取值范围为[0"," π/4],则点P的横坐标的取值范围为 .
解析:设点P的横坐标为x0, . ]
∵f'(x0)=lim┬(Δx"→" 0) ("(" x_0+Δx")" ^2+2"(" x_0+Δx")" +3"-(" x_0^2+2x_0+3")" )/Δx
=(lim)┬(Δx"→" 0) ("(" 2x_0+2")·" Δx+"(" Δx")" ^2)/Δx
=lim┬(Δx"→" 0)(Δx+2x0+2)=2x0+2,
∴曲线C在点P处的切线斜率为2x0+2.由已知得0≤2x0+2≤1,∴-1≤x0≤-1/2,∴点P的横坐标的取值范围为["-" 1",-" 1/2].
答案:["-" 1",-" 1/2]
10.若抛物线y=f(x)=4x2上的点P到直线y=4x-5的距离最短,求点P的坐标.
解由点P到直线y=4x-5的距离最短知,过点P的切线方程与直线y=4x-5平行.设P(x0,y0),则f'(x0)=lim┬(Δx"→" 0) Δy/Δx=(lim)┬(Δx"→" 0) (4"(" x_0+Δx")" ^2 "-" 4x_0^2)/Δx=lim┬(Δx"→" 0) (8x_0 "·" Δx+4"(" Δx")" ^2)/Δx=lim┬(Δx"→" 0)(8x0+4Δx)=8x0.由{■(8x_0=4"," @y_0=4x_0^2 "," )┤得{■(x_0=1/2 "," @y_0=1"," )┤故所求的点P的坐标为(1/2 "," 1).
★11.求过点(-1,-2)且与曲线y=f(x)=2x-x3相切的直线方程.
解设切点坐标为(x0,2x0-x_0^3),
f'(x0)=lim┬(Δx"→" 0) Δy/Δx=(lim)┬(Δx"→" 0) (2"(" x_0+Δx")-(" x_0+Δx")" ^3 "-" 2x_0+x_0^3)/Δx
=lim┬(Δx"→" 0)[2-3x_0^2-3x0Δx-(Δx)2]=2-3x_0^2.
则切线方程为y-2x0+x_0^3=(2-3x_0^2)(x-x0).
∵切线过点(-1,-2), 学 ]
∴-2-2x0+x_0^3=(2-3x_0^2)(-1-x0),
即2x_0^3+3x_0^2=0,解得x0=0或x0=-3/2.
故切点坐标为(0,0)或("-" 3/2 "," 3/8). Z