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tangent f(x)=sqrt(x^2+1),\at x=1

解答

切线

f (x) = x^{2} + 1, at x = 1

解答

y = \frac{1}{2} x + 2 - \frac{1}{2}

求解步骤

找到切点:

(1, 2)

计算

f (x) = x^{2} + 1

的斜率:

\frac{df ( x )}{d x} = \frac{x}{x ^{2} + 1}

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}{2}

找到斜率 m=

\frac{1}{2}

且穿过

(1, 2)

的直线:

y = \frac{1}{2} x + 2 - \frac{1}{2}

y = \frac{1}{2} x + 2 - \frac{1}{2}

作图

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