Question

given the function below

$f(x)=\sqrt3{-80 x^{3}+144}$

find the equation of the tangent line to the graph of the function at $x = 1$. answer in $m x+b$ form.

$l(x)=$

use the tangent line to approximate $f(1.1)$.

$l(1.1)=$

compute the actual value of $f(1.1)$. what is the error between the function value and the linear approximation?
answer as a positive value only.

$|$error$| \approx$ (approximate to at least 5 decimal places.)

question help: video

Explanation:

Step1: 求\(f(1)\)

将\(x = 1\)代入\(f(x)=\sqrt[3]{-80x^{3}+144}\)，得\(f(1)=\sqrt[3]{-80\times1^{3}+144}=\sqrt[3]{64}=4\)。

Step2: 求\(f^{\prime}(x)\)

根据复合函数求导法则\((u^n)^\prime=nu^{n - 1}u^\prime\)，令\(u=-80x^{3}+144\)，\(n=\frac{1}{3}\)，则\(f^{\prime}(x)=\frac{1}{3}(-80x^{3}+144)^{-\frac{2}{3}}\times(-240x^{2})=\frac{-80x^{2}}{(-80x^{3}+144)^{\frac{2}{3}}}\)。

Step3: 求\(f^{\prime}(1)\)

将\(x = 1\)代入\(f^{\prime}(x)\)，得\(f^{\prime}(1)=\frac{-80\times1^{2}}{(-80\times1^{3}+144)^{\frac{2}{3}}}=\frac{-80}{64^{\frac{2}{3}}}=\frac{-80}{16}=-5\)。

Step4: 求切线方程\(L(x)\)

根据点斜式\(y - y_0 = m(x - x_0)\)（这里\(x_0 = 1\)，\(y_0 = 4\)，\(m=-5\)），\(L(x)-4=-5(x - 1)\)，整理得\(L(x)=-5x + 9\)。

Step5: 求\(L(1.1)\)

将\(x = 1.1\)代入\(L(x)=-5x + 9\)，得\(L(1.1)=-5\times1.1 + 9=3.5\)。

Step6: 求\(f(1.1)\)

\(f(1.1)=\sqrt[3]{-80\times(1.1)^{3}+144}=\sqrt[3]{-80\times1.331 + 144}=\sqrt[3]{-106.48+144}=\sqrt[3]{37.52}\approx3.3476\)。

Step7: 求\(\vert error\vert\)

\(\vert error\vert=\vert f(1.1)-L(1.1)\vert=\vert3.3476 - 3.5\vert=0.1524\)。

Answer:

\(L(x)=-5x + 9\)；\(L(1.1)=3.5\)；\(\vert error\vert\approx0.1524\)。