### Video Transcript

For the given function π« of π‘ equals two π to the power of π‘ minus one π’ plus cos of π‘π π£ minus the natural log of π‘ π€, evaluate π« of one.

In this question, weβve been given 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. Weβre being asked to evaluate π« of one. In other words, we need to let π‘ be equal to one. So we replace π‘ with one in each part of our function.

So the component for π’ becomes two π to the power of one minus one. The component of π£ is cos of one π. And the component for π€ is negative the natural log of one. Now of course, two π to the power of one minus one is two π to the power of zero. And cos of one π is equal to cos of π. So we have two π to the power of zero π’ plus cos of π π£ minus the natural log of one π€.

But of course, anything to the power of zero is one. And so the component of π’ here is two. We know cos of π to be equal to negative one. So the component for π£ is negative one. And we can write π« of one as two π’ minus π£. But we also know that the natural log of one is zero. So π« of one is two π’ minus one π£ plus zero π€, which is simply two π’ minus π£.