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

Evaluate the rate of change of π(π₯) = β(3π₯ β 1) at π₯ = 7.

06:12

### Video Transcript

Evaluate the rate of change of the function π of π₯ is equal to the square root of three π₯ minus one at π₯ is equal to seven.

The question gives us a function π of π₯. It wants us to find the rate of change of this function π of π₯ when π₯ is equal to seven. We recall we denote the rate of change of a function π of π₯ at π₯ is equal to π by π prime evaluated at π. And we say that this is equal to the limit as β approaches zero of π evaluated at π plus β minus π evaluated at π divided by β if this limit exists.

In our case, we want to find the derivative when π₯ is equal to seven. So weβll set π equal to seven. So weβll set π equal to seven in our definition of the rate of change. This gives us the limit we need to evaluate. To help us evaluate this limit, letβs first find an expression for π evaluated at seven plus β. π evaluated at seven plus β means we substitute π₯ is equal to seven plus β into our definition for π of π₯. Doing this, we get the square root of three times seven plus β minus one.

We can then simplify this expression by distributing three over our parentheses. This gives us 21 plus three β. And remember, we then need to subtract one. This gives us 20 plus three β. So weβve shown π evaluated at seven plus β is equal to the square root of 20 plus three β. We can do something similar for π evaluated at seven. We substitute π₯ is equal to seven into our definition of π of π₯. This gives us the square root of three times seven minus one. And three times seven minus one is equal to root 20. So π evaluated at seven is the square root of 20.

So weβre now ready to try and find the rate of change of our function π of π₯ at π₯ is equal to seven. From our definition, we have that π prime of seven is equal to the limit as β approaches zero of π evaluated at seven plus β minus π evaluated at seven divided by β if this limit exists.

Next, weβll substitute in our expressions for π evaluated at seven plus β and π evaluated at seven. This gives us the limit as β approaches zero of the square root of 20 plus three β minus the square root of 20 divided by β.

At this point, we might be tempted to try evaluating our limit by direct substitution. However, doing this, we end up with the square root of 20 minus the square root of 20 all divided by zero, which simplifies to give us the indeterminate form zero divided by zero. So weβre going to need a different method to evaluate this limit.

To help us evaluate this limit, we recall the algebraic manipulation which says if we have the sum or difference between two surds, we can manipulate this by multiplying by the conjugate. We recall the complex conjugate means instead of taking the difference between our two surds, we instead add the two surds. And weβll multiply both the numerator and the denominator by this conjugate so that our limit remains the same.

So we want to multiply these two fractions inside of our limit. Weβll start with the numerator and do this by using the FOIL method. We multiply the first term in each of our numerators. We can see that this gives us the square root of 20 plus three β multiplied by the square root of 20 plus three β. In other words, we get the square root of 20 plus three β multiplied by itself. We can write this as it being squared. And the square root of a number squared is just equal to itself. So our first term is just 20 plus three β.

Next, the FOIL method tells us to multiply the outer terms. This gives us the square root of 20 plus three β times the square root of 20. Next, we need to multiply our inner terms. This gives us negative root 20 times root 20 plus three β. Finally, we want to multiply the last two terms. This gives us negative root 20 multiplied by root 20, which we can simplify to give us negative 20.

So we found the following expression for our numerator. We could find an expression for our denominator. However, letβs first simplify our numerator. First, we see we have 20 minus 20. So these cancel. Next, we can cancel our center two terms. The square root of 20 multiplied by root 20 plus three β minus root 20 multiplied by root 20 plus three β cancel to give us zero. So our numerator just simplifies to give us three β.

Now, we can finally multiply the two terms in our denominator. So we now have the limit as β approaches zero of three β divided by β times root 20 plus three β plus root 20. And we can simplify this further. Our limit is as β is approaching zero. So β is not equal to zero. So our limit will be the same if we just cancel the shared factor of β in our numerator and our denominator. So we get the limit of the following quotient.

And now we can attempt to evaluate this by direct substitution. Substituting β is equal to zero into our limit, we get three divided by the square root of 20 plus three times zero plus the square root of 20. And we can just calculate this. First, three multiplied by zero is equal to zero. So our denominator is now the square root of 20 plus the square root of 20. This is equal to two times root 20. So our limit evaluates to give us three divided by two root 20.

And we could leave our answer like this. However, we could also simplify this by multiplying the numerator and the denominator by the square root of 20. Doing this, we get three root 20 divided by two times 20. And of course, two times 20 is equal to 40.

The last bit of manipulation weβll do is weβll notice that 20 is equal to four times five. This tells us the square root of 20 is equal to the square root of four times five. Then, using our laws of exponents, weβll write this as the square root of four times the square root of five, which is of course just equal to two root five. So rewriting root 20 as two root five, we get three times two root five divided by 40. And weβll divide both our numerator and our denominator through by two. This gives us three root five divided by 20, which is our final answer.

Therefore, weβve shown if π of π₯ is equal to the square root of three π₯ minus one, then the rate of change of π of π₯ at π₯ is equal to seven is equal to three root five divided by 20.