PART 12：例題-反函數的微分 【92交大財金所】

Let  \(f(x) = {x^5} + 2{x^3} + 3x - 7\) ，Evaluate  \(f(1)\) and  \({\left( {{f^{ - 1}}} \right)^{\prime \prime }}( - 1)\)

解析:這一題不可能先找出 \({f^{ - 1}}(x)\) ，而且題目已經提示先要我們計算 \(f(1)\)

SOL:

(1) 首先計算 \(f(1) = 1 + 2 + 3 - 7 =  - 1\) ，提示了 \({f^{ - 1}}( - 1) = 1\)

(2) \({\left[ {{f^{ - 1}}(x)} \right]^\prime } = \frac{1}{{f'({f^{ - 1}}(x))}}\) ，再次微分

　 \({\left[ {{f^{ - 1}}(x)} \right]^{\prime \prime }} = \frac{{ - f''({f^{ - 1}}(x))({f^{ - 1}}(x))'}}{{{{\left[ {f'({f^{ - 1}}(x))} \right]}^2}}} = \frac{{ - f''({f^{ - 1}}(x))}}{{{{\left[ {f'({f^{ - 1}}(x))} \right]}^3}}}\)

(3) 令 \(x =  - 1\)， \({\left[ {{f^{ - 1}}} \right]^{\prime \prime }}( - 1) = \frac{{ - f''({f^{ - 1}}( - 1))}}{{{{\left[ {f'({f^{ - 1}}( - 1))} \right]}^3}}}\)  \( = \frac{{ - f''(1)}}{{{{\left[ {f'(1)} \right]}^3}}}\)

(4) \(f'(x) = 5{x^4} + 6{x^2} + 3\;\; \Rightarrow \;\;f'(1) = 14\)

　 \(f''(x) = 20{x^3} + 12x\;\; \Rightarrow \;\;f''(1) = 32\) 代入(3)

(5)  \({\left[ {{f^{ - 1}}} \right]^{\prime \prime }}( - 1)\) \( = \frac{{ - f''(1)}}{{{{\left[ {f'(1)} \right]}^3}}} = \frac{{ - 32}}{{{{14}^3}}} = \frac{{ - 4}}{{7 \cdot 7 \cdot 7}} = \frac{{ - 4}}{{343}}\)