Question

given the function below
$f(x)=\sqrt3{-108 x^{3}+324}$
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.)
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Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=\sqrt[3]{- 108x^{3}+324}\), we can rewrite it as \(f(x)=(-108x^{3}+324)^{\frac{1}{3}}\).
Using the chain - rule \((u^n)^\prime = nu^{n - 1}u^\prime\), where \(u=-108x^{3}+324\) and \(n = \frac{1}{3}\).
First, \(u^\prime=-324x^{2}\).
Then \(f^\prime(x)=\frac{1}{3}(-108x^{3}+324)^{-\frac{2}{3}}\times(-324x^{2})=\frac{- 108x^{2}}{(-108x^{3}+324)^{\frac{2}{3}}}\).

Step2: Evaluate \(f(1)\) and \(f^\prime(1)\)

For \(f(1)\):
\(f(1)=\sqrt[3]{-108\times1^{3}+324}=\sqrt[3]{216}=6\).
For \(f^\prime(1)\):
\(f^\prime(1)=\frac{-108\times1^{2}}{(-108\times1^{3}+324)^{\frac{2}{3}}}=\frac{-108}{216^{\frac{2}{3}}}\).
Since \(216 = 6^{3}\), then \(216^{\frac{2}{3}}=(6^{3})^{\frac{2}{3}} = 36\).
So \(f^\prime(1)=\frac{-108}{36}=-3\).

Step3: Find the equation of the tangent line \(L(x)\)

The equation of the tangent line in the form \(L(x)=mx + b\) (where \(m = f^\prime(a)\) and \(b=f(a)-f^\prime(a)\times a\), here \(a = 1\))
Using the point - slope form \(y - y_0=m(x - x_0)\) (\((x_0,y_0)=(1,6)\) and \(m=-3\))
\(y-6=-3(x - 1)\)
\(y-6=-3x + 3\)
\(L(x)=-3x+9\).

Step4: Approximate \(f(1.1)\) using \(L(x)\)

Substitute \(x = 1.1\) into \(L(x)\):
\(L(1.1)=-3\times1.1 + 9=-3.3+9=5.7\).

Step5: Compute the actual value of \(f(1.1)\)

\(f(1.1)=\sqrt[3]{-108\times(1.1)^{3}+324}=\sqrt[3]{-108\times1.331 + 324}=\sqrt[3]{-143.748+324}=\sqrt[3]{180.252}\approx5.648\).

Step6: Compute the error

\(\vert\text{error}\vert=\vert f(1.1)-L(1.1)\vert=\vert5.648 - 5.7\vert=0.052\).

Answer:

\(L(x)=-3x + 9\); \(L(1.1)=5.7\); \(f(1.1)\approx5.648\); \(\vert\text{error}\vert\approx0.052\)