Question

1. let (f(x)=3+\frac{5}{x}). a. determine (f(x)). (f(x)=) b. determine the slope of the tangent line at (x = 3). slope (=f(3)=)

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=3 + \frac{5}{x}$ as $f(x)=3+5x^{- 1}$.

Step2: Apply 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 $5x^{-1}$, using the power - rule with $a = 5$ and $n=-1$, we have $(5x^{-1})^\prime=5\times(-1)x^{-1 - 1}=-5x^{-2}$. So, $f^\prime(x)=- \frac{5}{x^{2}}$.

Step3: Find the slope of the tangent line at $x = 3$

Substitute $x = 3$ into $f^\prime(x)$. We get $f^\prime(3)=-\frac{5}{3^{2}}=-\frac{5}{9}$.

Answer:

a. $f^\prime(x)=-\frac{5}{x^{2}}$
b. $f^\prime(3)=-\frac{5}{9}$