# Video: Vector-Valued Functions

For the given function π«(π‘) = (π‘Β² β 1)π’ + π‘π£, find π«(π‘ β 1 ) β π«(π‘).

02:07

### 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. Itβs a function whose range is a vector or set of vectors, and whose domain is a subset of the real numbers. Now, weβre looking to find π« of π‘ minus one minus π« of π‘. Well, we know what π« of π‘ is, so letβs find what π« of π‘ minus one is. Weβre 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£.

Weβre 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, thatβs negative π‘, and the inner terms, thatβs 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‘. Thatβs π‘ squared minus two π‘ π’ plus π‘ minus one π£ β remember, thatβs 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, weβre 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, weβve established that π« of π‘ minus one minus π« of π‘ is negative two π‘ plus one π’ minus π£.