Answer:
![f(d)=\left\{ \begin{matrix} 11 & \text{for} & 0< d\leq 4 \\-3+3.5d & \text{for}& d > 4 \end](https://tex.z-dn.net/?f=f%28d%29%3D%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%2011%20%26%20%5Ctext%7Bfor%7D%20%26%200%3C%20d%5Cleq%204%20%20%5C%5C-3%2B3.5d%20%26%20%5Ctext%7Bfor%7D%26%20%20d%20%3E%204%20%20%5Cend)
Step-by-step explanation:
Cost of first 4 kilometers ride = 11 pesos.
So, for the first 4 km, the fair is constant i.e.
for 1 km the fare is 11 pesos,
for 2 km the fair is also 11 pesos,
similarly, for 3 km of 4 km, the fare is 11 pesos.
Hence, for the distance
, the fair function,
![f(d)=11\cdots(i)](https://tex.z-dn.net/?f=f%28d%29%3D11%5Ccdots%28i%29)
After 4 km, there is an increment of $ 3.50 for each kilometer.
So, the fare function up to 5 kilometers,
![f(d)=11+3.5=14.5](https://tex.z-dn.net/?f=f%28d%29%3D11%2B3.5%3D14.5)
So, the fare function up to 6 kilometers, i.e for the distance
,
![f(d)=11+2\times3.5=18](https://tex.z-dn.net/?f=f%28d%29%3D11%2B2%5Ctimes3.5%3D18)
This can be arranged as the fare function up to 6 km,
![f(d)=11+(6-4)\times3.5=18](https://tex.z-dn.net/?f=f%28d%29%3D11%2B%286-4%29%5Ctimes3.5%3D18)
Similarly, the fare function up to
kilometer
,
![f(d)=11+(d-4)\times3.5](https://tex.z-dn.net/?f=f%28d%29%3D11%2B%28d-4%29%5Ctimes3.5)
![\Rightarrrow f(d)=-3+3.5d\cdots(ii)](https://tex.z-dn.net/?f=%5CRightarrrow%20f%28d%29%3D-3%2B3.5d%5Ccdots%28ii%29)
Hence, from equations (i) and (ii),
![f(d)=\left\{ \begin{matrix} 11 & \text{for}\; 0< d \leq 4 \\-3+3.5d & \text{for}\; d > 4 \end](https://tex.z-dn.net/?f=f%28d%29%3D%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%2011%20%26%20%5Ctext%7Bfor%7D%5C%3B%200%3C%20d%20%5Cleq%204%20%20%5C%5C-3%2B3.5d%20%26%20%5Ctext%7Bfor%7D%5C%3B%20%20d%20%3E%204%20%20%5Cend)