202101141227 Exercise 1 (Q10)

If $f(x)=2x^2$ , find expressions for:

1. $f(x+h)$ .

2. $f(x+h)-f(x)$ .

3. $\displaystyle{\frac{f(x+h)-f(x)}{h}}$ .

Solution.

1.

$\begin{aligned} f(x+h) & = 2(x+h)^2 \\ & = 2(x^2+2hx+h^2) \\ & = 2x^2+4hx+2h^2 \end{aligned}$

2.

$\begin{aligned} f(x+h)-f(x) & = 2(x+h)^2 - 2(x)^2 \\ & = 2(x^2+2hx+h^2) - 2x^2 \\ & = 4hx+2h^2 \\ \end{aligned}$

3.

$\begin{aligned} \frac{f(x+h)-f(x)}{h} & = \frac{2(x+h)^2-2(x)^2}{h} \\ & = \frac{2(x^2+2hx+h^2)-2x^2}{h} \\ & = \frac{4hx+2h^2}{h} \\ & = 4x+2h \\ \end{aligned}$

Remark.

The first step to derivation from first principles, is calculate the slope of tangent

$\displaystyle{\frac{f(x+h)-f(x)}{h}}$ ,

and put the limit $h\to 0$ , i.e.,

$\begin{aligned} f'(x) & =\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}x}} \\ & = \displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}} \end{aligned}$

a right-hand limit.

Afterword.

$\begin{aligned} f(x) & =2x^2 \\ f'(x) & = \frac{\mathrm{d}f}{\mathrm{d}x} \\ & = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & \stackrel{\textrm{(3)}}{=} \lim_{h\to 0} 4x+2h \\ & = 4x+2(0) \\ & = 4x \end{aligned}$

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