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108 學年度台綜大微積分 B 第 3 題

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108學年度 · 108台綜大微積分B · 第 3 題

題目

Problem

(3) (10 pts) Let f:(−δ,δ)→Rf:(-\delta,\delta)\to\mathbb{R} be a function so that f(0)=12f(0)=\dfrac{1}{2} for δ>0\delta>0. Assume that f(x)f(x) satisfies the equation

x2+(f(x))2=(2x2+2(f(x))2−x)2x^2+\bigl(f(x)\bigr)^2=\bigl(2x^2+2\bigl(f(x)\bigr)^2-x\bigr)^2

for all x∈(−δ,δ)x\in(-\delta,\delta). Suppose we know that ff is differentiable at 00. Compute the tangent line to the curve y=f(x)y=f(x) at the point (0,12)\left(0,\dfrac{1}{2}\right).

解答

解法一

令 y=f(x)y=f(x),並把右側括號記作

A=2x2+2y2−x.A=2x^2+2y^2-x.

對隱式關係 x2+y2=A2x^2+y^2=A^2 兩邊對 xx 微分:

2x+2y\frac{\mathrm{d}y}{\mathrm{d}x} =&\,2A\left(4x+4y\frac{\mathrm{d}y}{\mathrm{d}x}-1\right).

在 (0,12)\left(0,\frac12\right) 處,A=12A=\frac12,所以

\frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\frac{\mathrm{d}y}{\mathrm{d}x}-1, \qquad \frac{\mathrm{d}y}{\mathrm{d}x}=1.

切線經過 (0,12)\left(0,\frac12\right),斜率為 11,故為

y=x+12.\boxed{y=x+\frac12}.