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For j(x) = 4x − 2, find j of the quantity x plus h end quantity minus j of x all over h period


4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h end quantity all over h

4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h plus 1 end quantity all over h

4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h minus 1 end quantity all over h

the quantity x minus 2 end quantity times the quantity 4 to the power of h plus 1 end quantity all over h

PLEASE HELPFor jx 4x 2 find j of the quantity x plus h end quantity minus j of x all over h period 4 to the power of the quantity x minus 2 end quantity times t class=

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Step-by-step explanation:

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

                     [tex]\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}[/tex]

                                                            [tex]\\ \\ \\ \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}[/tex]

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