Question

PLEASE HELP

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

Answer 1

Answer: It is C

Explanation:

Got it right

Answer 2

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(\Delta y)/(\Delta x) \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle(f\left(x+h\right) \ - \ f\left(x\right))/(\left(x \ + \ h \right) \ - \ x) \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle(f\left(x + h\right) \ - \ f\left(x\right))/(h) \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle(4^(x+h-2) \ - \ 4^(x-2))/(h) \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle(4^(x-2+h) \ - \ 4^(x-2))/(h)`

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