Question

1. find $f(x)$ by determining $lim_{h \to 0}\frac{f(x + h)-f(x)}{h}$, given $f(x)=x^{2}-2x$.

Explanation:

Step1: Find \(f(x + h)\)

Substitute \(x+h\) into \(f(x)\):
\[f(x + h)=(x + h)^2-2(x + h)=x^{2}+2xh+h^{2}-2x - 2h\]

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

\[

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

\]

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

\[

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

\]

Step4: Find the limit as \(h

ightarrow0\)
\[

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

\]

Answer:

\(f'(x)=2x - 2\)