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Find the domain of the vector-valued function π of π‘ equals π‘ squared plus four π plus tan π‘ π plus two times the natural log of π‘ π.
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We recall that the domain of a function is all the values it can take, all the possible inputs.
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And we can find the domain of a vector-valued function by looking for the intersection of the domains of each of our component functions.
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So, weβre going to need to work out the domain of π‘ squared plus four, the domain of tan π‘, and the domain of two times the natural log of π‘.
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Weβll begin by looking at the function that represents the component for π.
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Thatβs π‘ squared plus four.
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π‘ squared plus four is a polynomial function.
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And we know that the domain of a polynomial function is all real numbers.
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So, we can say that π‘ must be a real number.
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And we found the domain of π‘ squared plus four.
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But what about the domain of tan of π‘?
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Letβs begin by recalling what the graph of π¦ equals tan of π₯ looks like.
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It looks a little something like this.
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Notice at π by two and then intervals of π radians, there is an asymptote.
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These are the values for which tan of π₯ is undefined.
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We can therefore say that the domain of tan π‘ is all values of π‘ not including π‘ is negative one π by two, one π by two, three π by two, and so one.
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Another way of saying this is π‘ cannot be equal to two π plus one multiples of π by two, where π takes an integer value.
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And now, we know the domain of tan of π‘.
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Finally, letβs find the domain of two times the natural log of π‘.
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We know that the natural log of π‘ cannot take values of π‘ less than or equal to zero.
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So, the domain of the natural log of π‘, and therefore two times the natural log of π‘, is π‘ is greater than zero.
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Remember, the domain of our vector-valued function is the intersection of these domains.
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π‘ is therefore a real number greater than zero which is not equal to two π plus one multiples of π by two for integer values of π.