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tangent f(x)= 1/(2sqrt(x)),\at x=64

解答

切线

f (x) = \frac{1}{2 x}, at x = 64

解答

y = - \frac{1}{2048} x + \frac{3}{32}

求解步骤

找到切点:

(64, \frac{1}{16})

计算

f (x) = \frac{1}{2 x}

的斜率:

\frac{df ( x )}{d x} = - \frac{1}{4 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{1}{2048}

找到斜率 m=

- \frac{1}{2048}

且穿过

(64, \frac{1}{16})

的直线:

y = - \frac{1}{2048} x + \frac{3}{32}

y = - \frac{1}{2048} x + \frac{3}{32}

作图

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