given the function below\nfind the equation of the tangent line to the graph of the function at x = 1…

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}\cdot 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}}\cdot(-324x^{2})=\frac{- 108x^{2}}{(-108x^{3}+324)^{\frac{2}{3}}}).

Step2: Evaluate (f(x)) and (f^\prime(x)) at (x = 1)

When (x = 1), (f(1)=\sqrt[3]{-108\times1^{3}+324}=\sqrt[3]{216}=6). (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 (y=mx + b) (where (m=f^\prime(a)) and (b=f(a)-f^\prime(a)\cdot 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)), we get (y-6=-3(x - 1)). Expanding, (y=-3x+3 + 6), so (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}\approx5.64877).

Step6: Compute the error

(\vert error\vert=\vert f(1.1)-L(1.1)\vert=\vert5.64877 - 5.7\vert=0.05123).

Answer:

(L(x)=-3x + 9); (L(1.1)=5.7); (f(1.1)\approx5.64877); (\vert error\vert\approx0.05123)