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Find the instantaneous rate of change of π of π₯ is equal to the square root of π₯ at π₯ equals π₯ one, which is greater than zero.
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Remember, the instantaneous rate of change of a function π of π₯ at a point π₯ equals π is found by taking the limit as β approaches zero of the average rate of change function.
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Thatβs the limit as β approaches zero of π of π plus β minus π of π all over β.
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In this case, we know that π of π₯ is equal to the square root of π₯, and we want to find the instantaneous rate of change at π₯ equals π₯ one.
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So, weβll let π be equal to π₯ one.
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Letβs substitute what we know into our formula.
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We want to compute the limit as β approaches zero of π of π₯ one plus β minus π of π₯ one all over β.
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We need to find the limit as β approaches zero of the square root of π₯ one plus β minus the square root of π₯ one all over β.
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Now, we canβt do this with direct substitution.
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If we do, we end up dividing by zero and we know that to be undefined.
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And so instead, we multiply the numerator and denominator of the function by the conjugate of the numerator, by the square root of π₯ plus one plus β plus the square root of π₯ one.
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On the denominator, we simply have β times the square root of π₯ one plus β plus the square root of π₯ one.
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Then on the numerator, we have the square root of π₯ one plus β times the square root of π₯ one plus β, which is simply π₯ one plus β.
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Then, we multiply the square root of π₯ one plus β by the square root of π₯, and negative the square root of π₯ one times the square root of π₯ one plus β.
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When we find their sum, we get zero.
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So, all thatβs left to do is to multiply negative the square root of π₯ one by the square root of π₯ one.
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And we simply get negative π₯ one.
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π₯ one minus π₯ one is zero.
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And then, we divide through by β.
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And so, this becomes the limit as β approaches zero of one over the square root of π₯ one plus β plus the square root of π₯ one.
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And we can now evaluate this as β approaches zero.
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Weβre left with one over the square root of π₯ one plus the square root of π₯ one, which is one over two times the square root of π₯ one.
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The instantaneous rate of change function of π of π₯ is equal to the square root of π₯ is therefore one over two times the square root of π₯ one.