Question Video: Vector-Valued Functions | Nagwa Question Video: Vector-Valued Functions | Nagwa

# Question Video: Vector-Valued Functions Mathematics • Higher Education

For the given function π«(π‘) = 2π^(π‘ β 1) π’ + cos (π‘π)π£ β ln (π‘)π€, evaluate π«(1).

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### 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, weβve been given 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. Weβre 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£.

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