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2020年9月28日 星期一

國立臺灣大學一百零九學年度轉學生入學考試試題詳解

※注意：請於試卷上「非選擇題作答區」標明大題及小題題號，並依序作答。

Any device with computer algebra system is prohibited during the exam.

PART 1: Fill in the blanks.

  • Only answers will be graded.
  • Each answer must be clearly labeled on the answer sheet.
  • points are assigned to each blank.
    1. lim　(1)　, where \left[x\right] is the greatest integer which is less than or equal to x.
    2. \displaystyle\lim_{x\to0}\left(\cos ax\right)^{\textstyle\frac1{1-\cos bx}}=　(2)　, where a,b are constants and ab\neq0.
    3. \displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac1{\displaystyle2n\sqrt{1-\left(\frac i{2n}\right)^2}}=　(3)　.
    2. 訣竅運用夾擠定理與自然指數的定義處理第一小題；第二小題需先使用換底公式，再搭配 L'Hôpital 法則求解；第三小題可轉換為 Riemann 和後取極限寫為定積分計算之。
    解法
    1. 注意到 x-1<\left[x\right]\leq x，因此有

      \displaystyle\left[\left(1+\frac3x\right)^{\frac x3}\right]^3\left(1+\frac3x\right)^{-1}\leq\left(1+\frac3x\right)^{\left[x\right]}\leq\left[\left(1+\frac3x\right)^{\frac x3}\right]^3.

      又由自然指數的定義可注意到

      \displaystyle\lim_{x\to\infty}\left[\left(1+\frac3x\right)^{\frac x3}\right]^3\left(1+\frac3x\right)^{-1}=e^3=\lim_{x\to\infty}\left[\left(1+\frac3x\right)^{\frac x3}\right]^3.

      從而由夾擠定理可知所求亦為 e^3。
    2. 由換底公式與指數函數的連續性並搭配 L'Hôpital 法則可知

      \displaystyle\lim_{x\to0}\left(\cos ax\right)^{\textstyle\frac1{1-\cos bx}}=\exp\left(\lim_{x\to0}\frac{\ln\left(\cos ax\right)}{1-\cos bx}\right)=\exp\left(\lim_{x\to0}\frac{-a\sin ax/\cos ax}{b\sin bx}\right)=\exp\left(-\frac{a^2}{b^2}\right)=e^{-\frac{a^2}{b^2}}.

    3. 所求可寫為 \displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac1{\displaystyle\sqrt{4-\left(\frac{i}n\right)^2}}。可將此視為函數 \displaystyle f\left(x\right)=\frac1{\sqrt{4-x^2}} 在區間 \displaystyle\left[0,1\right] 上作 n 等分割的 Riemann 和，故其極限值可化為定積分並計算如下

      \displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac1{\displaystyle2n\sqrt{1-\left(\frac{i}{2n}\right)^2}}=\int_0^1\frac{dx}{\sqrt{4-x^2}}=\left.\sin^{-1}\left(\frac x2\right)\right|_0^1=\sin^{-1}\frac12=\frac\pi6.


    1. \displaystyle f\left(x\right)=\left[\ln\left(1+x^2\right)\right]^x, \displaystyle\frac d{dx}f\left(x\right)=　(4)　.
    2. x^3-y^2+y^3=x. At \left(x,y\right)=\left(0,1\right). \displaystyle\frac{d^2y}{dx^2}=　(5)　.
    3. \displaystyle f\left(x,y,z\right)=\int_z^{x^y}e^{\sqrt t}dt. \nabla f=　(6)　.
    4. \displaystyle f\left(x,y\right)=\frac{\sin\left(x^2y\right)}{x^2+y^2} for \left(x,y\right)\neq\left(0,0\right) and f\left(0,0\right)=0. The directional derivative of f along {\bf u}=\left(\cos\theta,\sin\theta\right) at \left(0,0\right) is 　(7)　.
    3. 訣竅運用換底公式後使用連鎖律等微分法則可求解第一小題；使用隱函數微分求導函數的值；搭配微分積分定理並留意梯度的定義計算；運用方向導數的定義式計算求解，此題由於分段定義的情況故不可逕自化為梯度與方向的內積，需確認函數的可微性等資訊才可直接使用（而本題屬於不可使用的情形）。
    解法
    1. 透過換底可將 f 表達為 f\left(x\right)=\exp\left[x\ln\left(\ln\left(1+x^2\right)\right)\right]，如此求導有

      \begin{aligned}f'\left(x\right)&=\exp\left[x\ln\left(\ln\left(1+x^2\right)\right)\right]\cdot\left[\ln\left(\ln\left(1+x^2\right)\right)+x\cdot\frac1{\ln\left(1+x^2\right)}\cdot\frac1{1+x^2}\cdot2x\right]\\&=\left(\ln\left(1+x^2\right)\right)^x\left(\frac{2x^2}{\left(1+x^2\right)\ln\left(1+x^2\right)}+\ln\left(\ln\left(1+x^2\right)\right)\right).\end{aligned}

    2. 運用隱函數微分求導可得

      \displaystyle3x^2-2y\frac{dy}{dx}+3y^2\frac{dy}{dx}=1.

      取 \left(x,y\right)=\left(0,1\right) 代入可得 \displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(0,1\right)}=1。又進一步求導有

      \displaystyle6x-2\left(\frac{dy}{dx}\right)^2-2y\frac{d^2y}{dx^2}+6y\left(\frac{dy}{dx}\right)^2+3y^2\frac{d^2y}{dx^2}=0.

      因此由 \left(x,y\right)=\left(0,1\right) 與 \displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(0,1\right)}=1 可推得 \displaystyle\left.\frac{d^2y}{dx^2}\right|_{\left(x,y\right)=\left(0,1\right)}=-4。
    3. 按梯度的定義可知

      \nabla f\left(x,y,z\right)=\left(f_x\left(x,y,z\right),f_y\left(x,y,z\right),f_z\left(x,y,z\right)\right)=\left(yx^{y-1}e^{x^{y/2}},x^ye^{x^{y/2}}\ln x,-e^{\sqrt z}\right).

    4. 由方向導數的定義可知所求為

      \displaystyle D_{\bf u}f\left(0,0\right)=\lim_{h\to0}\frac{f\left(h\cos\theta,h\sin\theta\right)-f\left(0,0\right)}h=\lim_{h\to0}\frac{\sin\left(h^3\cos^2\theta\sin\theta\right)}{h^3}=\cos^2\theta\sin\theta.


  3. f\left(x\right)=\begin{cases}\left|x\right|-1,&\text{for}~-1\leq x\leq3\\\left(x-6\right)^2-4,&\text{for}~3<x\leq10.\end{cases}. \displaystyle g\left(x\right)=\int_0^xf\left(t\right)dt, for -1\leq x\leq10.
    1. Choose correct statement(s) about g\left(x\right): 　(8)　
      1. g\left(x\right) is discontinuous at x=3.
      2. g\left(x\right) is not differentiable at x=3.
      3. g\left(x\right) is differentiable at x=0 and g'\left(0\right)=-1.
      4. g\left(x\right)<0 for -1\leq x<0.
    2. The local minimum values of g\left(x\right) occur at x=　(9)　.
    3. The x-coordinates of points of inflection for y=g\left(x\right) are 　(10)　.
    4. 訣竅由函數圖形與積分的意義的概念可回答關於函數 g 的特性，隨後由球導的狀況回答局部極值的位置，並進一步求其反曲點。
    解法
    1. 由於 f 在 x=3 處左右極限均存在但不相等，從而不定積分 g 在 x=3 處連續但不可導，而在可導處的求導結果與 f 相同，即有： g'\left(0\right)=f\left(0\right)=-1 以及

      \displaystyle g\left(x\right)=\int_0^xf\left(t\right)dt=\int_x^0\left(1-\left|t\right|\right)dt=\int_x^0\left(1+t\right)dt=-x-\frac{x^2}2=-\frac12\left(x+1\right)^2+\frac12>0.

      即應選ii. 與 iii.
    2. 由於當 x\neq3 時有 g'\left(x\right)=f\left(x\right)，從而解 g'\left(x\right)=0 可知有 x=\pm1、x=4 與 x=8。又不可導的位置有 x=3，而邊界為 x=-1 與 x=10。逐步考量斜率的遞增遞減的區域可發現：函數 g 在 \left(-1,1\right) 處遞減，隨後在 \left(1,3\right) 上遞增，又於 \left(3,4\right) 上遞增，接著 \left(4,8\right) 遞減，最終於 \left(8,10\right) 上遞增。綜述可知在 x=1、x=8 處有局部極值。
    3. 在除了 x=0 與 x=3 處有 g''\left(x\right)=\begin{cases}-1,&\text{for}~-1<x<0,\\1,&\text{for}~0<x<3,\\2x-12,&\text{for}~3<x<10.\end{cases}。從而可知在 x=0、x=3 與 x=6 等處有凹向性的改變。

  4. A vertical fence 2 m high is located 1 m away from a wall. The length of the shortest ladder that can extend from the wall over the fence to a point on the ground is 　(11)　.
    5. 訣竅依據題意設定參數並表達出階梯的長度為該參數的函數，並透過微分求其極值。
    解法設此階梯與籬笆的距離為 x，那麼按題設可知與高牆的距離為 x+1，從而使用相似三角形可知該階梯的長度為 L\left(x\right)=\left(1+x^{-1}\right)\sqrt{x^2+4}，其中 x>0。對此求導有

    \displaystyle L'\left(x\right)=-x^{-2}\sqrt{x^2+4}+(1+x)\left(x^2+4\right)^{-1/2}=x^{-2}\left(x^2+4\right)^{-1/2}\left(x^3-4\right).

    因此解 L'\left(x\right)=0 有 x=2^{2/3}，且容易注意到 x\in\left(0,2^{2/3}\right) 有 L'\left(x\right)<0 而 x\in\left(2^{2/3},\infty\right) 有 L'\left(x\right)>0。故在 x=2^{2/3} 達到最小值。此時階梯長度的最小值為 \left(1+2^{2/3}\right)^{3/2}。

  5. Compute the following integrals:
    1. \displaystyle\int_0^{\infty}\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}\,dx=　(12)　.
    2. \displaystyle\int\frac{dx}{\sqrt{x^2-6x}}=　(13)　.
    3. \displaystyle\int\sin^{-1}\left(\sqrt x\right)dx=　(14)　.
    6. 訣竅由有理函數的積分技巧、三角代換法與一般的變數代換與分部積分法答題。
    解法
    1. 首先注意到被積分函數可改寫如下

      \displaystyle\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}=-\frac{1/5}{x-1}+\frac{x/5}{x^2+4}+\frac{11/5}{x^2+4},

      其中 x=1 處為瑕點。因此將給定的瑕積分分拆為兩段如下：

      \displaystyle\int_0^\infty\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx=\int_0^{1^-}\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx+\int_{1^+}^\infty\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx.

      然而這兩者皆發散：

      \displaystyle\begin{aligned}\int_0^{1^-}\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx&=\int_0^{1^-}\left(\frac{1/5}{1-x}+\frac{x/5}{x^2+4}+\frac{11/5}{x^2+4}\right)dx\\&=\left.\left(-\frac15\ln\left(1-x\right)+\frac1{10}\ln\left(x^2+4\right)+\frac{11}{10}\tan^{-1}\frac x2\right)\right|_0^{1^-}=\infty.\end{aligned}

      類似地也有 \displaystyle\int_{1^+}^\infty\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx=-\infty，故給定的瑕積分發散。
    2. 配方有 x^2-6x=\left(x-3\right)^2-9。令 x-3=3\sec\theta，那麼有 dx=3\sec\theta\tan\theta\,d\theta，如此所求的不定積分可改寫並計算如下

      \displaystyle\begin{aligned}\int\frac{dx}{\sqrt{x^2-6x}}&=\int\frac{3\sec\theta\tan\theta \,d\theta}{3\tan\theta}=\int\sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\\&=\ln\left|\frac{x-3+\sqrt{x^2-6x}}3\right|+C=\ln\left|x-3+\sqrt{x^2-6x}\right|+C'.\end{aligned}

      此處 C'=C-\ln3。
    3. 令 u=\sin^{-1}\left(\sqrt x\right)，如此有 x=\sin^2u 與 dx=2\sin u\cos u\,du=\sin2u\,du。據此所求的不定積分為

      \displaystyle\begin{aligned}\int\sin^{-1}\left(\sqrt x\right)dx&=\int u\sin2u\,du=-\frac{u\cos2u}2+\frac12\int\cos2u\,du\\&=-\frac{2u\cos2u-\sin2u}4+C=-\frac{u\left(\cos^2u-\sin^2u\right)-\sin u\cos u}2+C\\&=-\frac{\left(1-2x\right)\sin^{-1}\left(\sqrt x\right)-\sqrt{x-x^2}}2+C.\end{aligned}


    1. \displaystyle\iint_Dy\,dA=　(15)　, where D is the region in the first quadrant that lies between the circles x^2+y^2=4 and \left(x-1\right)^2+y^2=1.
    2. The volume of the solid R lying below the plane z=3-2y and above the paraboloid z=x^2+y^2 is 　(16)　.
    7. 訣竅對於第一小題可運用極座標變換處理即可；第二小題則先求出其交會處，並利用底乘以高的概念列式求其體積。
    解法
    1. 令 x=r\cos\theta、y=r\sin\theta，根據積分區域可得變數範圍為 2\cos\theta\leq r\leq2、0\leq\theta\leq\pi/2。如此所求之重積分可改寫並計算如下

      \displaystyle\begin{aligned}\iint_Dy\,dA&=\int_0^{\pi/2}\int_{2\cos\theta}^2r\sin\theta\cdot r\,dr\,d\theta=\int_0^{\pi/2}\left.\frac{r^3\sin\theta}3\right|_{2\cos\theta}^2d\theta\\&=\frac83\int_0^{\pi/2}\left(\sin\theta-\sin\theta\cos^3\theta\right)d\theta=\left.\frac83\left(\frac{\cos^4\theta}4-\cos\theta\right)\right|_0^{\frac\pi2}=2.\end{aligned}

    2. 首先先考慮兩曲面的交線由 x^2+y^2=3-2y 所決定，此即 x^2+\left(y+1\right)^2=4，如此我們設定 D=\left\{\left(x,y\right)\in\mathbb R^2\,:\,x^2+\left(y+1\right)^2\leq4\right\}。那麼所求的體積可列式如下

      \displaystyle V=\iint_D\left[\left(3-2y\right)-\left(x^2+y^2\right)\right]dA=\iint_D\left[4-x^2-\left(y+1\right)^2\right]dA.

      那麼令 x=r\cos\theta、y+1=r\sin\theta，如此變數範圍為 0\leq r\leq2、0\leq\theta\leq2\pi，則所求的體積可改寫並計算如下

      \displaystyle V=\int_0^{2\pi}\int_0^2\left(4-r^2\right)r\,dr\,d\theta=2\pi\left.\left(2r^2-\frac{r^4}4\right)\right|_0^2=8\pi.


    1. C is a smooth curve in the upper half plane going from \left(1,1\right) to \left(0,\sqrt2\right). \displaystyle{\bf F}\left(x,y\right)=\frac{x-3y}{x^2+y^2}{\bf i}+\frac{3x+y}{x^2+y^2}{\bf j}. Then, \displaystyle\int_C{\bf F}\cdot d{\bf r}=　(17)　.
    2. S is the part of the sphere x^2+y^2+z^2=9 that lies above the cone z=\sqrt{x^2+y^2} with upward orientation. \displaystyle{\bf F}\left(x,y,z\right)=\frac{x{\bf i}+y{\bf j}+z{\bf k}}{\left(x^2+y^2+z^2\right)^2}. Then, \displaystyle\iint_S{\bf F}\cdot d{\bf S}=　(18)　.
    8. 訣竅試圖找出位能函數並使用線積分基本定理即可；將取面參數化後直接計算即可。
    解法
    1. 若函數 f 滿足 \nabla f={\bf F}，那麼由 \displaystyle f_x\left(x,y\right)=\frac{x-3y}{x^2+y^2} 可知

      \displaystyle f\left(x,y\right)=\frac12\ln\left(x^2+y^2\right)-3\tan^{-1}\frac xy+g\left(y\right).

      如此對 y 求導便有

      \displaystyle f_y\left(x,y\right)=\frac y{x^2+y^2}+3\cdot\frac x{y^2+x^2}+g'\left(y\right)=\frac{3x+y}{x^2+y^2}.

      故 g'\left(y\right)=0，從而 g 為常數函數。如此取位能函數為

      \displaystyle f\left(x,y\right)=\frac12\ln\left(x^2+y^2\right)-3\tan^{-1}\frac xy+C.

      因此所求為

      \displaystyle\int_C{\bf F}\cdot d{\bf r}=f\left(0,\sqrt2\right)-f\left(1,1\right)=\frac{3\pi}4.

    2. 將曲面參數化為 {\bf r}\left(\theta,\phi\right)=\left(3\cos\theta\sin\phi,3\sin\theta\sin\phi,3\cos\phi\right)，其中參數範圍 0\leq\theta\leq2\pi、0\leq\phi\leq\pi/4，那麼

      \displaystyle\begin{aligned}\left({\bf r}_\theta\times{\bf r}_\phi\right)d\theta\,d\phi&=\left(\left(-3\sin\theta\sin\phi,3\cos\theta\sin\phi,0\right)\times\left(3\cos\theta\cos\phi,3\sin\theta\cos\phi,-3\sin\phi\right)\right)d\theta\,d\phi\\&=\left(-9\cos\theta\sin^2\phi,-9\sin\theta\sin^2\phi,-9\sin\phi\cos\phi\right)d\theta\,d\phi\\&=-9\sin\phi\left(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi\right)d\theta\,d\phi\end{aligned}

      表指向球心的有向面元，如此所求的曲面積分為

      \displaystyle\begin{aligned}\iint_S{\bf F}\cdot d{\bf S}&=\int_0^{\pi/4}\int_0^{2\pi}\frac{\left(3\cos\theta\sin\phi,3\sin\theta\sin\phi,3\cos\phi\right)}{9^2}\cdot9\sin\phi\left(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi\right)d\theta\,d\phi\\&=\frac13\int_0^{\pi/4}\int_0^{2\pi}\sin\phi\,d\theta\,d\phi=\frac13\cdot2\pi\left(1-\frac{\sqrt2}2\right)=\frac{\left(2-\sqrt2\right)\pi}3.\end{aligned}


    1. \displaystyle\sum_{n=0}^{\infty}\frac{3^n}{\left(2n\right)!}=　(19)　.
    2. The Maclaurin series (the Taylor series at a=0) for \cos^{-1}x is 　(20)　.
    9. 訣竅利用經典的 Taylor 級數求給定的級數值並運用廣義二項式定理求冪級數。
    解法
    1. 首先觀察知道

      \displaystyle\frac{e^x+e^{-x}}2=\frac12\left(\sum_{n=0}^\infty\frac{x^n}{n!}+\sum_{n=0}^\infty\frac{\left(-1\right)^nx^n}{n!}\right)=\sum_{n=0}^\infty\frac{x^{2n}}{\left(2n\right)!}.

      因此取 x=\sqrt3 可得

      \displaystyle\sum_{n=0}^\infty\frac{3^n}{\left(2n\right)!}=\frac{e^{\sqrt3}+e^{-\sqrt3}}2=\cosh\left(\sqrt3\right).

    2. 首先回憶起 \displaystyle\cos^{-1}x=\frac\pi2-\sin^{-1}x，又

      \displaystyle\begin{aligned}\sin^{-1}x&=\int_0^x\frac{dt}{\sqrt{1-t^2}}=\int_0^x\left(1-t^2\right)^{-1/2}dt=\int_0^x\sum_{n=0}^{\infty}{-1/2\choose n}\left(-t^2\right)^ndt\\&=\int_0^x\sum_{n=0}^\infty\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2}t^{2n}dt=\sum_{n=0}^\infty\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2\left(2n+1\right)}x^{2n+1}.\end{aligned}

      因此所求的冪級數為

      \displaystyle\cos^{-1}x=\frac\pi2-\sum_{n=0}^\infty\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2\left(2n+1\right)}x^{2n+1}.

PART 2:

  • Solve the following problems. You need to write down complete arguments.
  • 10 points are assigned to each problem.
  1. Find the equation of the curve on the xy-plane that passes through \left(3,1\right) such that for any point P on the curve, the midpoint of the tangent line at P that lies in the first quadrant is P itself.
    2. 訣竅設定參數曲線並按題意考慮切線在兩軸交點的中點列出方程式。
    解法設所求的曲線參數式為 \left(x\left(t\right),y\left(t\right)\right)，那麼若 P=\left(x_0,y_0\right)=\left(x\left(t_0\right),y\left(t_0\right)\right)，則在該點的切線方程式為

    \displaystyle y-y_0=\frac{y'\left(t_0\right)}{x'\left(t_0\right)}\left(x-x_0\right).

    此時兩軸上的交點為

    \displaystyle\left(0,y_0-x_0\frac{y'\left(t_0\right)}{x'\left(t_0\right)}\right),\qquad\left(x_0-y_0\frac{x'\left(t_0\right)}{y'\left(t_0\right)},0\right).

    由中點關係得關係式

    \displaystyle x_0-y_0\frac{x'\left(t_0\right)}{y'\left(t_0\right)}=2x_0,\quad y_0-x_0\frac{y'\left(t_0\right)}{x'\left(t_0\right)}=2y_0.

    由於 t_0 的任意性可知 \displaystyle\frac{x'\left(t\right)}{y'\left(t\right)}=-\frac{x\left(t\right)}{y\left(t\right)}，或寫為 \displaystyle\frac{dx}x+\frac{dy}y=0，因此積分得 \ln\left|xy\right|=C，從而有 xy=e^C=C'。再者又因通過 \left(3,1\right)，故此曲線為 xy=3。

  2. f\left(x,y\right) is a differentiable function. On the curve \displaystyle\tan^{-1}\left(xy\right)+y=2+\frac\pi4, f obtains local maximum at \displaystyle\left(x,y\right)=\left(\frac12,2\right). Suppose that \displaystyle f_x\left(\frac12,2\right)=2.
    1. Find \displaystyle\nabla f\left(\frac12,2\right).
    2. Assume that on another curve \displaystyle\tan^{-1}\left(xy\right)+y=1.9+\frac\pi4, f obtains local maximum at \left(x_0,y_0\right) which is near \displaystyle\left(\frac12,2\right). Use linear approximation to estimate \displaystyle f\left(x_0,y_0\right)-f\left(\frac12,2\right).
    3. 訣竅由 Lagrange 乘子法的概念推理。
    解法
    1. 設函數 \displaystyle g\left(x,y\right)=\tan^{-1}\left(xy\right)+y-2-\frac\pi4，則限制條件為 g\left(x,y\right)=0。如此由 Lagrange 乘子法的概念可知極值發生的位置滿足條件

      \displaystyle\left(2,f_y\left(\frac12,2\right)\right)=\left(f_x\left(\frac12,2\right),f_y\left(\frac12,2\right)\right)=\nabla f\left(\frac12,2\right)\parallel\nabla g\left(\frac12,2\right)=\left(g_x\left(\frac12,2\right),g_y\left(\frac12,2\right)\right).

      又計算函數 g 的梯度有

      \displaystyle\nabla g\left(x,y\right)=\left(\frac y{1+\left(xy\right)^2},\frac x{1+\left(xy\right)^2}+1\right).

      取 \displaystyle \left(x,y\right)=\left(\frac12,2\right) 代入便有 \displaystyle\nabla g\left(\frac12,2\right)=\left(1,\frac54\right)，故知 \displaystyle f_y\left(\frac12,2\right)=\frac52，從而 \displaystyle\nabla f\left(\frac12,2\right)=\left(2,\frac52\right)。
    2. 運用一階逼近可推知

      \displaystyle\begin{aligned}f\left(x_0,y_0\right)-f\left(\frac12,2\right)&\approx\nabla f\left(\frac12,2\right)\cdot\left(x_0-\frac12,y_0-2\right)=2\nabla g\left(\frac12,2\right)\cdot\left(x_0-\frac12,y_0-2\right)\\&\approx2\left[g\left(x_0,y_0\right)-g\left(\frac12,2\right)\right]=2\cdot\left(-0.1\right)=-0.2.\end{aligned}

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