given the function below\n$f(x)=\\sqrt3{45 x^{3}-18}$\nfind the equation of the tangent line to the graph of…

given the function below\n$f(x)=\\sqrt3{45 x^{3}-18}$\nfind the equation of the tangent line to the graph of the function at $x = 1$. answer in $m x+b$ form.\n$l(x)=\\square$\nuse the tangent line to approximate $f(1.1)$.\n$l(1.1)=\\square$\ncompute the actual value of $f(1.1)$. what is the error between the function value and the linear approximation?\nanswer as a positive value only.\n$|$error$| \\approx \\square$\n(approximate to at least 5 decimal places.)
Answer
Explanation:
Step1: Find the derivative of (f(x))
Given (f(x)=\sqrt[3]{45x^{3}-18}=(45x^{3}-18)^{\frac{1}{3}}). Using the chain rule ((u^{n})^\prime = nu^{n - 1}u^\prime), where (u = 45x^{3}-18) and (n=\frac{1}{3}). (u^\prime=135x^{2}). So (f^\prime(x)=\frac{1}{3}(45x^{3}-18)^{-\frac{2}{3}}\times135x^{2}=\frac{45x^{2}}{(45x^{3}-18)^{\frac{2}{3}}}).
Step2: Evaluate (f(1)) and (f^\prime(1))
- (f(1)=\sqrt[3]{45\times1^{3}-18}=\sqrt[3]{27} = 3).
- (f^\prime(1)=\frac{45\times1^{2}}{(45\times1^{3}-18)^{\frac{2}{3}}}=\frac{45}{27^{\frac{2}{3}}}=\frac{45}{9}=5).
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)a), here (a = 1)). (m = 5), (b=f(1)-f^\prime(1)\times1=3 - 5\times1=-2). So (L(x)=5x-2).
Step4: Approximate (f(1.1)) using (L(x))
Substitute (x = 1.1) into (L(x)): (L(1.1)=5\times1.1-2=5.5 - 2=3.5).
Step5: Compute the actual value of (f(1.1))
(f(1.1)=\sqrt[3]{45\times(1.1)^{3}-18}=\sqrt[3]{45\times1.331-18}=\sqrt[3]{59.895 - 18}=\sqrt[3]{41.895}\approx3.47476).
Step6: Compute the error
(\vert\text{error}\vert=\vert f(1.1)-L(1.1)\vert=\vert3.47476 - 3.5\vert=0.02524).
Answer:
- (L(x)=5x - 2)
- (L(1.1)=3.5)
- (f(1.1)\approx3.47476)
- (\vert\text{error}\vert\approx0.02524)