Answered

Get the most out of your questions with IDNLearn.com's extensive resources. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.

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

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=

Sagot :

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]

View image Seanhewwzx
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.