Question

use the formula $f(x)=lim_{z \to x}\frac{f(z)-f(x)}{z - x}$ to find the derivative of the following function.
$f(x)=2x^{2}-x + 4$
$f(x)=4x - 1$

Explanation:

Step1: Find f(z)

Substitute z into f(x): $f(z)=2z^{2}-z + 4$

Step2: Calculate f(z)-f(x)

\[

$$\begin{align*} f(z)-f(x)&=(2z^{2}-z + 4)-(2x^{2}-x + 4)\\ &=2z^{2}-z + 4 - 2x^{2}+x - 4\\ &=2(z^{2}-x^{2})-(z - x) \end{align*}$$

\]

Step3: Simplify $\frac{f(z)-f(x)}{z - x}$

\[

$$\begin{align*} \frac{f(z)-f(x)}{z - x}&=\frac{2(z^{2}-x^{2})-(z - x)}{z - x}\\ &=\frac{2(z - x)(z + x)-(z - x)}{z - x}\\ &=\frac{(z - x)(2(z + x)-1)}{z - x}\\ &=2(z + x)-1 \end{align*}$$

\]

Step4: Find the limit as $z

ightarrow x$
\[

$$\begin{align*} f'(x)&=\lim_{z ightarrow x}\frac{f(z)-f(x)}{z - x}\\ &=\lim_{z ightarrow x}(2(z + x)-1)\\ &=2(x + x)-1\\ &=4x-1 \end{align*}$$

\]

Answer:

$4x - 1$