Question

Between x = 2 and x = 3, which function has an average rate of change that is double that of as y = 1/2^-x + 1

Answer 1

Given :

Give function is :

$y= \frac{1}{2^{x} } + 1$

To Find :

A function which have double  rate of change that of given function between x = 2 and x = 3 , = ?

Solution :

∴For the given function , it is give that the avg rate of change between x = 2 and x = 3 is given as :

Δ  = $\frac{y(3)- y(2)}{3-2}$

    = $\frac{(\frac{1}{2^{3} }+1)-(\frac{1}{2^{2} }+1)}{3-2}$

    $= \frac{\frac{1}{8}+1-\frac{1}{4}-1}{1}$

    $=\frac{-1}{8}$

Now after making this change double  , it becomes :

$= \frac{-1}{4}$

So we have to find a function which has an avg rate of change :

$= \frac{-1}{4}$

∴$y'(3)-y'(2)=\frac{-1}{4}$

Hence y' can be =$\frac{-x}{4}$

As, $\frac{-3}{4}-(\frac{-2}{4})= \frac{-3}{4}+\frac{2}{4}= \frac{-1}{4}$

So the required function is $\frac{-x}{4}$