Let $f(x)=6x^2$
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 $f^{\prime }(x)=12x$.

Step 3 ：At $x=a=1$, $f^{\prime }(1)=12$.

Step 4 ：The linearization $L(x)$ of $f$ at $a=1$ is given by $L(x)=f(a)+f^{\prime }(a)(x-a)$.

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 $0.06$.