Question

1. (5 points) let f(x)=2x² - 4x + 6. find the derivative f(x) by using the definition of the derivative of f as a limit of a difference quotient, and use algebraic calculations to compute the limit.

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = f(x)$ is $f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=2x^{2}-4x + 6$, first find $f(x + h)$.
\[

$$\begin{align*} f(x + h)&=2(x + h)^{2}-4(x + h)+6\\ &=2(x^{2}+2xh+h^{2})-4x-4h + 6\\ &=2x^{2}+4xh+2h^{2}-4x-4h + 6 \end{align*}$$

\]

Step2: Calculate $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh+2h^{2}-4x-4h + 6)-(2x^{2}-4x + 6)\\ &=2x^{2}+4xh+2h^{2}-4x-4h + 6-2x^{2}+4x - 6\\ &=4xh+2h^{2}-4h \end{align*}$$

\]

Step3: Calculate $\frac{f(x + h)-f(x)}{h}$

\[

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

\]

Step4: Compute the limit as $h

ightarrow0$
\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(4x+2h - 4)\\ &=4x-4 \end{align*}$$

\]

Answer:

$f'(x)=4x - 4$