Question Video: Evaluating the Rate of Change of a Radical Function at a Point | Nagwa Question Video: Evaluating the Rate of Change of a Radical Function at a Point | Nagwa

# Question Video: Evaluating the Rate of Change of a Radical Function at a Point Mathematics • Second Year of Secondary School

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Evaluate the rate of change of π(π₯) = β(6π₯ + 7) at π₯ = 3.

03:12

### Video Transcript

Evaluate the rate of change of π of π₯ equals the square root of six π₯ plus seven at π₯ equals three.

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. In this question, π of π₯ is equal to the square root of six π₯ plus seven. And weβre looking to find the rate of change at π₯ equals three. So weβre going to let π be equal to three. The rate of change is the derivative of our function evaluated at π₯ equals three. Using our earlier definition, we see that itβs the limit as β approaches zero of π of three plus β minus π of three all over β.

Our job is going to be to work out π of three plus β and π of three. To find π of three plus β, we replace π₯ in our original expression for the function with three plus β. So itβs the square root of six times three plus β plus seven. Distributing the parentheses and we get 18 plus six β plus seven, which simplifies to six β plus 25. So π of three plus β is the square root of six β plus 25. We repeat this process for π of three. This time, itβs the square root of six times three plus seven, which is root 25 or simply five.

Substituting this back into our original definition for π prime of three and we find that itβs now equal to the limit as β approaches zero of the square root of six β plus 25 minus five over β. Well, weβre not quite ready to perform direct substitution. If we did, weβd be dividing by zero, which we know to be undefined. So instead, weβre going to need to do something a little bit clever to manipulate our expression. We begin by writing the square root of six β plus 25 as six β plus 25 to the power of one-half. Weβre then going to multiply the numerator and denominator of our limit by the conjugate of the numerator. So thatβs six β plus 25 to the power of one-half plus five.

Letβs distribute the numerator. We begin by multiplying the first term in each expression. Now, six β plus 25 to the power of one-half times itself. Well, thatβs simply six β plus 25. We multiply six β plus 25 to the power of one-half by five. And then, we multiply negative five by six β plus 25 to the power of one-half. 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. Finally, we multiply negative five by five and we get negative 25. For now, weβll leave the denominator as shown. Letβs clear some space for the next step.

We noticed that 25 minus 25 is zero. So our numerator becomes six β. And then, we spot that we can simplify by dividing through by β. And in fact, weβre now ready to perform direct substitution. Weβre going to replace β in our limit with zero. Then, our denominator becomes six times zero plus 25 to the power of one-half or 25 to the power of one-half. Well, 25 to the power of one-half is the square root of 25, which is five. So π prime of three is six over five plus five, which is, of course, six tenths. That simplifies to three-fifths. The rate of change of our function at π₯ equals three is, therefore, three-fifths.

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