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Evaluate the rate of change of π of π₯ equals the square root of six π₯ plus seven at π₯ equals three.
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Remember, the definition for the rate of change of a function or its derivative at a point π₯ equals π is the limit as β approaches zero of π of π plus β minus π of π over β.
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In this question, π of π₯ is equal to the square root of six π₯ plus seven.
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And weβre looking to find the rate of change at π₯ equals three.
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So weβre going to let π be equal to three.
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The rate of change is the derivative of our function evaluated at π₯ equals three.
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Using our earlier definition, we see that itβs the limit as β approaches zero of π of three plus β minus π of three all over β.
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Our job is going to be to work out π of three plus β and π of three.
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To find π of three plus β, we replace π₯ in our original expression for the function with three plus β.
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So itβs the square root of six times three plus β plus seven.
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Distributing the parentheses and we get 18 plus six β plus seven, which simplifies to six β plus 25.
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So π of three plus β is the square root of six β plus 25.
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We repeat this process for π of three.
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This time, itβs the square root of six times three plus seven, which is root 25 or simply five.
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Substituting this back into our original definition for π prime of three and we find that itβs now equal to the limit as β approaches zero of the square root of six β plus 25 minus five over β.
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Well, weβre not quite ready to perform direct substitution.
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If we did, weβd be dividing by zero, which we know to be undefined.
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So instead, weβre going to need to do something a little bit clever to manipulate our expression.
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We begin by writing the square root of six β plus 25 as six β plus 25 to the power of one-half.
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Weβre then going to multiply the numerator and denominator of our limit by the conjugate of the numerator.
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So thatβs six β plus 25 to the power of one-half plus five.
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Letβs distribute the numerator.
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We begin by multiplying the first term in each expression.
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Now, six β plus 25 to the power of one-half times itself.
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Well, thatβs simply six β plus 25.
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We multiply six β plus 25 to the power of one-half by five.
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And then, we multiply negative five by six β plus 25 to the power of one-half.
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And we end up with five lots of six β plus 25 to the power of one-half minus five lots of six β plus 25 to the power of one-half, which is, of course, zero.
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Finally, we multiply negative five by five and we get negative 25.
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For now, weβll leave the denominator as shown.
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Letβs clear some space for the next step.
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We noticed that 25 minus 25 is zero.
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So our numerator becomes six β.
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And then, we spot that we can simplify by dividing through by β.
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And in fact, weβre now ready to perform direct substitution.
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Weβre going to replace β in our limit with zero.
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Then, our denominator becomes six times zero plus 25 to the power of one-half or 25 to the power of one-half.
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Well, 25 to the power of one-half is the square root of 25, which is five.
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So π prime of three is six over five plus five, which is, of course, six tenths.
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That simplifies to three-fifths.
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The rate of change of our function at π₯ equals three is, therefore, three-fifths.