### Video Transcript

For the given function π« of π‘ equals π‘ squared minus one π’ plus π‘π£, find π« of π‘ minus one minus π« of π‘.

π« of π‘ is a vector-valued function. Itβs a function whose range is a vector or set of vectors, and whose domain is a subset of the real numbers. Now, weβre looking to find π« of π‘ minus one minus π« of π‘. Well, we know what π« of π‘ is, so letβs find what π« of π‘ minus one is. Weβre going to replace every instance of π‘ in our vector-valued function with π‘ minus one. When we do, we find that π« of π‘ minus one is π‘ minus one squared minus one π’ plus π‘ minus one π£.

Weβre going to carefully distribute the parentheses here. π‘ minus one squared is π‘ minus one times π‘ minus one. We multiply the first term in each expression. And we get π‘ squared. We then multiply the outer terms, thatβs negative π‘, and the inner terms, thatβs another negative π‘, then the last terms, negative one times negative one is positive one. So, we find π‘ minus one squared is equal to π‘ squared minus two π‘ plus one. And so, π« of π‘ minus one is π‘ squared minus two π‘ plus one minus one π’ plus π‘ minus one π£. Of course, one minus one is zero. So, we get π‘ squared minus two π‘ π’ plus π‘ minus one π£.

This question is asking us to find π« of π‘ minus one minus π« of π‘. Thatβs π‘ squared minus two π‘ π’ plus π‘ minus one π£ β remember, thatβs just π« of π‘ minus one β minus π« of π‘, which is π‘ squared minus one π’ plus π‘π£. To find the difference between these two two-dimensional vectors, weβre going to simply subtract the individual components. So, the π’-component will be π‘ squared minus two π‘ minus π‘ squared minus one. Then, the π£-component is π‘ minus one minus π‘. Now, π‘ squared minus π‘ squared is zero, and π‘ minus π‘ is zero. We then have negative two π‘ minus negative one, which gives us negative two π‘ plus one as a component for π’. And we simply have negative one as the component for π£. And so, weβve established that π« of π‘ minus one minus π« of π‘ is negative two π‘ plus one π’ minus π£.