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Mathematics

1 answer:

ratelena [41]2 years ago

8 0

Do you have to show work? i can do all of this for you

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2 years ago

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kykrilka [37]

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3 years ago

PLEASE HELP

lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function $j\left(x\right) \ = \ 4^{x-2}$ , hence the average rate of change (slope of the blue line) between the $x$ and $x+h$ is

$\text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}$

$\\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}$

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1 year ago