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tangent f(x)=(x^2)/(x+5),\at x=3

解答

切线

f (x) = \frac{x ^{2}}{x + 5}, at x = 3

解答

y = \frac{39}{64} x - \frac{45}{64}

求解步骤

找到切点:

(3, \frac{9}{8})

计算

f (x) = \frac{x ^{2}}{x + 5}

的斜率:

\frac{df ( x )}{d x} = \frac{x ^{2} + 10 x}{( x + 5 ) ^{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{39}{64}

找到斜率 m=

\frac{39}{64}

且穿过

(3, \frac{9}{8})

的直线:

y = \frac{39}{64} x - \frac{45}{64}

y = \frac{39}{64} x - \frac{45}{64}

作图

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