Question

compute (f^{prime}(c)) for each of the following using the alternate form of the definition of the derivative.

1. (f(x)=3x + 1, c = 1)
2. (f(x)=x+\frac{4}{x}, c = 4)

Explanation:

Step1: Recall the alternate - form of the derivative

The alternate form of the derivative is $f^{\prime}(c)=\lim_{x
ightarrow c}\frac{f(x)-f(c)}{x - c}$.

Step2: For $f(x)=3x + 1$ and $c = 1$

First, find $f(c)$:
When $c = 1$, $f(1)=3\times1+1=4$.
Then, $\frac{f(x)-f(1)}{x - 1}=\frac{(3x + 1)-4}{x - 1}=\frac{3x-3}{x - 1}=\frac{3(x - 1)}{x - 1}=3$ for $x
eq1$.
So, $f^{\prime}(1)=\lim_{x
ightarrow1}\frac{f(x)-f(1)}{x - 1}=\lim_{x
ightarrow1}3 = 3$.

Step3: For $f(x)=x+\frac{4}{x}$ and $c = 4$

First, find $f(c)$:
When $c = 4$, $f(4)=4+\frac{4}{4}=4 + 1=5$.
Then, $\frac{f(x)-f(4)}{x - 4}=\frac{(x+\frac{4}{x})-5}{x - 4}=\frac{x+\frac{4}{x}-5}{x - 4}=\frac{\frac{x^{2}+4-5x}{x}}{x - 4}=\frac{x^{2}-5x + 4}{x(x - 4)}$.
Factor the numerator: $x^{2}-5x + 4=(x - 1)(x - 4)$.
So, $\frac{(x - 1)(x - 4)}{x(x - 4)}=\frac{x - 1}{x}$ for $x
eq4$.
Then, $f^{\prime}(4)=\lim_{x
ightarrow4}\frac{f(x)-f(4)}{x - 4}=\lim_{x
ightarrow4}\frac{x - 1}{x}=\frac{4-1}{4}=\frac{3}{4}$.

Answer:

1. For $f(x)=3x + 1$ and $c = 1$, $f^{\prime}(1)=3$.
2. For $f(x)=x+\frac{4}{x}$ and $c = 4$, $f^{\prime}(4)=\frac{3}{4}$.