'The function $f(x)$ changes value when $x$ changes from $x_{0}$ to $x_{0}+d x$ \[ f(x)=2 x^{3}-2 x, x_{0}=1, d x=0.2 \] a. Find the change $\Delta f=f\left(x_{0}+d x\right)-f\left(x_{0}\right)$. b. Find the value of the estimate $d f=f^{\prime}\left(x_{0}\right) d x$ c. Find the approximation error $|\Delta f-d f|$.

Step 1 ：First, we need to find the value of the function $f(x)$ at $x_{0}$ and $x_{0}+dx$. Given $x_{0}=1$ and $dx=0.2$, we have $x_{0}+dx=1.2$.

Step 2 ：Substitute $x_{0}=1$ into the function $f(x)=2x^{3}-2x$, we get $f(x_{0})=2(1)^{3}-2(1)=0$.

Step 3 ：Substitute $x_{0}+dx=1.2$ into the function $f(x)$, we get $f(x_{0}+dx)=2(1.2)^{3}-2(1.2)=1.728$.

Step 4 ：The change in the function $f(x)$ when $x$ changes from $x_{0}$ to $x_{0}+dx$ is given by $\Delta f=f(x_{0}+dx)-f(x_{0})=1.728-0=1.728$.

Step 5 ：Next, we need to find the derivative of the function $f(x)$, $f'(x)=6x^{2}-2$.

Step 6 ：Substitute $x_{0}=1$ into the derivative $f'(x)$, we get $f'(x_{0})=6(1)^{2}-2=4$.

Step 7 ：The estimate of the change in the function $f(x)$ when $x$ changes from $x_{0}$ to $x_{0}+dx$ is given by $df=f'(x_{0})dx=4*0.2=0.8$.

Step 8 ：The approximation error is given by $|\Delta f-df|=|1.728-0.8|=0.928$.