如果题目是:
求过点(1,2)且在任一点(x,y)处切线斜率为x*(根号x)的曲线方程.
则可以:
∵f '(x) = k = x*根号x = x*x^(1/2) = x^(3/2)
∴f(x) = (2/5) * x^(5/2) + c
把(1,2)代入f(x) 得:
2 = (2/5) * 1 + c 解得:c = 8/5
所以f(x) = (2/5) * x^(5/2) + 8/5 (x>=0)
如果题目是:
求过点(1,2)且在任一点(x,y)处切线斜率为x*(根号x)的曲线方程.
则可以:
∵f '(x) = k = x*根号x = x*x^(1/2) = x^(3/2)
∴f(x) = (2/5) * x^(5/2) + c
把(1,2)代入f(x) 得:
2 = (2/5) * 1 + c 解得:c = 8/5
所以f(x) = (2/5) * x^(5/2) + 8/5 (x>=0)