Let $f(x)=6 x^{2}$ a) Find the linearization $L(x)$ of $f$ at $a=1$. b) Use the linearization to approximate $6(1.1)^{2}$. c) Find $6(1.1)^{2}$ using a calculator. d) What is the difference between the approximation and the actual value of $6(1.1)^{2}$.

Step 1 ：The function given is \(f(x) = 6x^{2}\).

Step 2 ：The derivative of \(f(x)\) is \(f'(x) = 12x\).

Step 3 ：At \(x = a = 1\), \(f'(1) = 12\).

Step 4 ：The linearization \(L(x)\) of \(f\) at \(a = 1\) is given by \(L(x) = f(a) + f'(a)(x - a)\).

Step 5 ：Substituting the values we get, \(L(x) = 6(1)^{2} + 12(x - 1)\).

Step 6 ：Simplifying, we get \(L(x) = 6 + 12(x - 1)\).

Step 7 ：So, the linearization \(L(x)\) of \(f\) at \(a = 1\) is \(L(x) = 6 + 12(x - 1)\).

Step 8 ：Now, we use the linearization to approximate \(6(1.1)^{2}\).

Step 9 ：Substitute \(x = 1.1\) into \(L(x)\), we get \(L(1.1) = 6 + 12(1.1 - 1)\).

Step 10 ：Simplifying, we get \(L(1.1) = 6 + 1.2 = 7.2\).

Step 11 ：So, the approximation of \(6(1.1)^{2}\) using the linearization is \(7.2\).

Step 12 ：Now, we find the actual value of \(6(1.1)^{2}\) using a calculator.

Step 13 ：The actual value of \(6(1.1)^{2}\) is \(7.26\).

Step 14 ：Finally, we find the difference between the approximation and the actual value of \(6(1.1)^{2}\).

Step 15 ：The difference is \(7.26 - 7.2 = 0.06\).

Step 16 ：So, the difference between the approximation and the actual value of \(6(1.1)^{2}\) is \(0.06\).