The given exponential function is
![f(x)=557(1.026)^x](https://img.qammunity.org/2023/formulas/mathematics/college/pnnk0b85az9ot2ayvtrhcv1tjp4rjopr74.png)
Where x is the number of years after 1968
f(x) is the population in millions
a) Substitute x by 0
![f(0)=557(1.026)^0](https://img.qammunity.org/2023/formulas/mathematics/college/gauzcfioejgm5klaqvv9hobq4pugixz342.png)
Since any number to the power of zero = 1, then
![\begin{gathered} (1.026)^0=1 \\ f(0)=557(1) \\ f(0)=557 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5pzar1ctk4he488mmmyofkj2baebdrwhjb.png)
The population in 1968 is 557 million
b) At year, 2000 we need to find the value of x
![\begin{gathered} x=2000-1968 \\ x=32 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ymsm3obef1rvupv4rn0j4ih90esua6abvm.png)
Now let us find f(32)
![\begin{gathered} f(32)=557(1.026)^(32) \\ f(32)=1266.399528 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7yttqjzkw8a9ptf2u7yiv6urehoo09h8ft.png)
Round it to the nearest whole number
Then the population in 2000 is 1266 million