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tangent f(x)= 5/(sqrt(x))

解答

切线

f (x) = \frac{5}{x}

解答

y = - \frac{5}{2 a _{0}^{\frac{3}{2}}} x + \frac{15}{2 a _{0}}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, \frac{5}{a _{0}})

计算

f (x) = \frac{5}{x}

的斜率:

\frac{df ( x )}{d x} = - \frac{5}{2 x ^{\frac{3}{2}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{5}{2 a _{0}^{\frac{3}{2}}}

找到斜率 m=

- \frac{5}{2 a _{0}^{\frac{3}{2}}}

且穿过

(a_{0}, \frac{5}{a _{0}})

的直线:

y = - \frac{5}{2 a _{0}^{\frac{3}{2}}} x + \frac{15}{2 a _{0}}

y = - \frac{5}{2 a _{0}^{\frac{3}{2}}} x + \frac{15}{2 a _{0}}

作图

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