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

Evaluate the rate of change of 𝑓(π‘₯) = √(6π‘₯ + 7) at π‘₯ = 3.

03:12

Video Transcript

Evaluate the rate of change of 𝑓 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‘Ž is the limit as β„Ž approaches zero of 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž over β„Ž. In this question, 𝑓 of π‘₯ is equal to the square root of six π‘₯ plus seven. And we’re looking to find the rate of change at π‘₯ equals three. So we’re going to let π‘Ž 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 it’s the limit as β„Ž approaches zero of 𝑓 of three plus β„Ž minus 𝑓 of three all over β„Ž.

Our job is going to be to work out 𝑓 of three plus β„Ž and 𝑓 of three. To find 𝑓 of three plus β„Ž, we replace π‘₯ in our original expression for the function with three plus β„Ž. So it’s the square root of six times three plus β„Ž plus seven. Distributing the parentheses and we get 18 plus six β„Ž plus seven, which simplifies to six β„Ž plus 25. So 𝑓 of three plus β„Ž is the square root of six β„Ž plus 25. We repeat this process for 𝑓 of three. This time, it’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 𝑓 prime of three and we find that it’s now equal to the limit as β„Ž approaches zero of the square root of six β„Ž plus 25 minus five over β„Ž. Well, we’re not quite ready to perform direct substitution. If we did, we’d be dividing by zero, which we know to be undefined. So instead, we’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„Ž plus 25 as six β„Ž plus 25 to the power of one-half. We’re then going to multiply the numerator and denominator of our limit by the conjugate of the numerator. So that’s six β„Ž plus 25 to the power of one-half plus five.

Let’s distribute the numerator. We begin by multiplying the first term in each expression. Now, six β„Ž plus 25 to the power of one-half times itself. Well, that’s simply six β„Ž plus 25. We multiply six β„Ž plus 25 to the power of one-half by five. And then, we multiply negative five by six β„Ž plus 25 to the power of one-half. And we end up with five lots of six β„Ž plus 25 to the power of one-half minus five lots of six β„Ž 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, we’ll leave the denominator as shown. Let’s clear some space for the next step.

We noticed that 25 minus 25 is zero. So our numerator becomes six β„Ž. And then, we spot that we can simplify by dividing through by β„Ž. And in fact, we’re now ready to perform direct substitution. We’re going to replace β„Ž 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 𝑓 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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