question given $f(x)=x^{3}-4x$, find the equation of the tangent line of $f$ at the point where $x = 2$.

Explanation:

Step1: Find the function value at x = 2

Substitute $x = 2$ into $f(x)=x^{3}-4x$. $f(2)=2^{3}-4\times2=8 - 8=0$

Step2: Find the derivative of the function

Use the power - rule. If $y = x^{n}$, then $y^\prime=nx^{n - 1}$. $f^\prime(x)=3x^{2}-4$

Step3: Find the slope of the tangent line at x = 2

Substitute $x = 2$ into $f^\prime(x)$. $f^\prime(2)=3\times2^{2}-4=12 - 4 = 8$

Step4: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(2,0)$ and $m = 8$. $y-0=8(x - 2)$ $y=8x-16$

Answer:

$y = 8x-16$