Given:
The function is
![f(x)=x^3-3x^2+5x-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/nsyzp4y6xla9zdi3ruosubz01ahx050m99.png)
To find:
The root of the function from the given possible roots.
Solution:
We have,
![f(x)=x^3-3x^2+5x-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/nsyzp4y6xla9zdi3ruosubz01ahx050m99.png)
At x=-3,
![f(-3)=(-3)^3-3(-3)^2+5(-3)-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/vm7jwlhon02egeid9ae3n1gy2gkyburotd.png)
![f(-3)=-27-27-15-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/n47xt2j9hw8tyhobd70vlzxps65epxvxnd.png)
![f(-3)=-84\\eq 0](https://img.qammunity.org/2021/formulas/mathematics/high-school/syozgyp33khw9sg3ha1jgn1bkgzx2otyge.png)
At x=-1,
![f(-1)=(-1)^3-3(-1)^2+5(-1)-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/r04a2umv68o1zx5sjwxu2s6sggdbp5o8h7.png)
![f(-1)=-1-3-5-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/q19fsn6yunapw6mu5s56mw0e21vzsdwi9t.png)
![f(-1)=-24\\eq 0](https://img.qammunity.org/2021/formulas/mathematics/high-school/icsdyuptm6a9gnguxa8tom0atey5sk29k0.png)
At x=1,
![f(1)=(1)^3-3(1)^2+5(1)-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/699f0y8rm1x8v0maab81c2116rn86ph8uz.png)
![f(1)=1-3+5-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/tdmx9r2mnsz7tkqqjholt77qoq7luh14tc.png)
![f(1)=-12\\eq 0](https://img.qammunity.org/2021/formulas/mathematics/high-school/yzu7iev0vmjqm3lxpmn4qnr105qk7me2is.png)
At x=3,
![f(3)=(3)^3-3(3)^2+5(3)-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/3swpm10kbqgzha18qm9ajg8fka9a44mc18.png)
![f(3)=27-27+15-15](https://img.qammunity.org/2021/formulas/mathematics/high-school/r6mwgr5qysjj8veh2i9x0icpdgjap4moof.png)
![f(3)=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/8bbzc7ad0lygu59cxufae665ma0uubtdk8.png)
Since, the value of given function is 0 at only x=3, therefore 3 is a root of given function.
Hence, the correct option is D.