2022-07-31

For each of the following function, evaluate this expression
$\frac{f\left(x+h\right)-f\left(x\right)}{h}$
1. $f\left(x\right)={x}^{2}-4x+5$
2. $g\left(x\right)={x}^{3}-2x+3$
3. $h\left(x\right)=\frac{1}{2x}$

Step 1
1) $\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\left(x+h{\right)}^{2}-4\left(x+h\right)+5-\left({x}^{2}-4x+5\right)}{h}$
$=\frac{\text{⧸}{x}^{2}+2xh+{h}^{2}-\text{⧸}4x-4h+\text{⧸}5-\text{⧸}{x}^{2}+\text{⧸}4x-\text{⧸}5}{h}$
$=\frac{h\left(2x+h-4\right)}{h}=2x+h-4$
Step 2
2) $\frac{g\left(x+h\right)-g\left(x\right)}{h}=\frac{\left(x+h{\right)}^{3}-2\left(x+h\right)+3-\left({x}^{3}-2x+3\right)}{h}$
$=\frac{{x}^{3}+3{x}^{2}h+3x{h}^{2}+{h}^{3}-2x-2h+3-{x}^{3}+2x-3}{h}$
$=\frac{h\left(3{x}^{2}+3xh+{h}^{2}-2\right)}{h}$
$=3{x}^{2}+3xh+{h}^{2}-2$
Step 3
3) $\frac{h\left(x+h\right)-h\left(x\right)}{h}=\frac{\frac{1}{2\left(x+h\right)}-\frac{1}{2x}}{h}$
$=\frac{2x-2\left(x+h\right)}{4xh\left(x+h\right)}$
$=\frac{-2h}{4xh\left(x+h\right)}=\frac{-1}{2x\left(x+h\right)}$

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