$\bf g(x)=\cfrac{1}{x^2}~\hspace{5em}\cfrac{g(x+h)-g(x)}{h}\implies \cfrac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} \\\\\\ \textit{using the LCD of }(x+h)^2(x^2)\qquad \cfrac{\frac{x^2-(x+h)^2}{(x+h)^2(x^2)}}{h}\implies \cfrac{x^2-(x+h)^2}{h(x+h)^2(x^2)} \\\\\\ \cfrac{x^2-(x^2+2xh+h^2)}{h(x+h)^2(x^2)}\implies \cfrac{\underline{x^2-x^2}-2xh-h^2}{h(x+h)^2(x^2)}\implies \cfrac{-2xh-h^2}{h(x+h)^2(x^2)}$

$\bf \cfrac{\underline{h}(-2x-h)}{\underline{h}(x+h)^2(x^2)}\implies \cfrac{-2x-h}{(x+h)^2(x^2)}\implies \cfrac{-2x-h}{(x^2+2xh+h^2)(x^2)} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{-2x-h}{x^4+2x^3h+x^2h^2}~\hfill$

4 0

3 years ago

Given y=5x−3 with the domain D: {−2, −1, 0, 1, 2}, fill in the missing values of the table below.

spayn [35]

Answer:

see below

Step-by-step explanation:

y = 5x-3

Let x = -2 y = 5(-2) -3 = -10 -3= -13

Let x = -1 y = 5(-1) -3 = -5-3 = -8

Let x = 0 y = 5(0) -3 = 0-3 = -3

Let x = 1 y = 5(1) -3 = 5-3 = 2

Let x = 2 y = 5(2) -3 =10-3 = 7

7 0

2 years ago