Given:
Consider the given function is:
![f(x)=5^x+1](https://tex.z-dn.net/?f=f%28x%29%3D5%5Ex%2B1)
To find:
The average rate of change between x = 0 and x = 4.
Solution:
The average rate of change of a function f(x) over the interval [a,b] is:
![m=\dfrac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
We have,
![f(x)=5^x+1](https://tex.z-dn.net/?f=f%28x%29%3D5%5Ex%2B1)
At
,
![f(0)=5^0+1](https://tex.z-dn.net/?f=f%280%29%3D5%5E0%2B1)
![f(0)=1+1](https://tex.z-dn.net/?f=f%280%29%3D1%2B1)
![f(0)=2](https://tex.z-dn.net/?f=f%280%29%3D2)
At
,
![f(4)=5^4+1](https://tex.z-dn.net/?f=f%284%29%3D5%5E4%2B1)
![f(4)=625+1](https://tex.z-dn.net/?f=f%284%29%3D625%2B1)
![f(4)=626](https://tex.z-dn.net/?f=f%284%29%3D626)
Now, the average rate of change between x = 0 and x = 4 is:
![m=\dfrac{f(4)-f(0)}{4-0}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bf%284%29-f%280%29%7D%7B4-0%7D)
![m=\dfrac{626-2}{4}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B626-2%7D%7B4%7D)
![m=\dfrac{624}{4}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B624%7D%7B4%7D)
![m=156](https://tex.z-dn.net/?f=m%3D156)
Hence, the average rate of change between x = 0 and x = 4 is 156.
Answer:
you're awesome
Step-by-step explanation:
tysmmm
<span>48m = 16(3m)
80n = 16(5n)
so
</span><span>48m - 80n = 16(3m - 5n)</span>