Question

assignment 4: problem 6 (1 point) let $f(x)=3+\frac{2}{x}+\frac{6}{x^{2}}$. find $f(x)$. $f(x)=square$ find $f(2)$. $f(2)=square$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=3 + \frac{2}{x}+\frac{6}{x^{2}}$ as $f(x)=3 + 2x^{-1}+6x^{-2}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
For the constant term $3$, its derivative is $0$ since the derivative of a constant $C$ is $0$.
For the term $2x^{-1}$, its derivative is $2\times(-1)x^{-1 - 1}=-2x^{-2}$.
For the term $6x^{-2}$, its derivative is $6\times(-2)x^{-2 - 1}=-12x^{-3}$.
So, $f^\prime(x)=0-2x^{-2}-12x^{-3}=-\frac{2}{x^{2}}-\frac{12}{x^{3}}$.

Step3: Find $f^\prime(2)$

Substitute $x = 2$ into $f^\prime(x)$.
$f^\prime(2)=-\frac{2}{2^{2}}-\frac{12}{2^{3}}=-\frac{2}{4}-\frac{12}{8}=-\frac{1}{2}-\frac{3}{2}=-2$.

Answer:

$f^\prime(x)=-\frac{2}{x^{2}}-\frac{12}{x^{3}}$
$f^\prime(2)=-2$