Video Transcript
Evaluate the instantaneous rate of change of 𝑓 of 𝑥 is equal to the square root of three 𝑥 minus one at 𝑥 equals seven.
This question gives us a function of 𝑥, and we are asked to find the instantaneous rate of change of this function 𝑓 of 𝑥 when 𝑥 is equal to seven. We recall that we denote the rate of change of a function 𝑓 of 𝑥 at 𝑥 is equal to 𝑎 by 𝑓 prime evaluated at 𝑎. And we say that this is equal to the limit as ℎ approaches zero of 𝑓 evaluated at 𝑎 plus ℎ minus 𝑓 evaluated at 𝑎 all divided by ℎ if this limit exists. In our case, we want to find the derivative when 𝑥 is equal to seven. We will therefore set 𝑎 equal to seven. This gives us the limit we need to evaluate as shown.
We will begin by finding an expression for 𝑓 evaluated at seven plus ℎ. To do this, we substitute 𝑥 is equal to seven plus ℎ into our definition for 𝑓 of 𝑥. This gives us the square root of three multiplied by seven plus ℎ minus one. Simplifying the expression underneath the square root, we have 21 plus three ℎ minus one. 𝑓 evaluated at seven plus ℎ is therefore equal to the square root of 20 plus three ℎ. We can repeat this process for 𝑓 evaluated at seven. Substituting 𝑥 equals seven into our definition of 𝑓 of 𝑥, we have the square root of three multiplied by seven minus one. This is equal to the square root of 20. So 𝑓 evaluated at seven is the square root of 20.
We are now in a position to try and find the rate of change of our function 𝑓 of 𝑥 at 𝑥 is equal to seven. From our definition, we have 𝑓 prime of seven is equal to the limit as ℎ approaches zero of the square root of 20 plus three ℎ minus the square root of 20 all divided by ℎ. At this point, we might be tempted to try evaluating our limit by direct substitution. However, when 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. We therefore need a different method to evaluate this limit.
To help us evaluate this limit, we recall the algebraic manipulation which says that if we have the sum or difference between two radicals or surds, we can manipulate this by multiplying by the conjugate. We recall the complex conjugate means instead of taking the difference between our two radicals, we instead add the two radicals. And we’ll multiply both the numerator and the denominator by this conjugate so that our limit remains the same. We want to multiply these two fractions inside of our limit. We can start with the numerator and do this using the FOIL method. Multiplying the first terms, we have the square root of 20 plus three ℎ multiplied by the square root of 20 plus three ℎ. This is equal to the square root of 20 plus three ℎ squared. And since the square root of a number squared is just equal to itself, our first term is just 20 plus three ℎ.
Next, we want to multiply the outer terms. This gives us the square root of 20 plus three ℎ multiplied by the square root of 20. Multiplying the inner terms gives us negative root 20 plus three ℎ multiplied by root 20. We note that these two terms cancel. Finally, we multiply the last terms negative root 20 multiplied by root 20, which is equal to negative 20. 20 minus 20 is equal to zero, so the numerator simplifies to three ℎ. On the denominator, we have ℎ multiplied by root 20 plus three ℎ plus root 20. This means that the expression simplifies to the limit as ℎ approaches zero of three ℎ over ℎ multiplied by root 20 plus three ℎ plus root 20.
We can simplify this one stage further. Our limit is as ℎ is approaching zero. Therefore, ℎ is not equal to zero. Our limit will therefore be the same if we just cancel the shared factor of ℎ in our numerator and denominator. And we end up with a limit of the following quotient. We can now attempt to evaluate this by direct substitution. Substituting ℎ is equal to zero into our limit, we have three over the square root of 20 plus three multiplied by zero plus the square root of 20. As three multiplied by zero is zero, this simplifies to three over root 20 plus root 20, which is equal to three over two root 20. And we could just leave our answer like this.
However, we could also simplify by multiplying the numerator and the denominator by the square root of 20. Doing this gives us three root 20 divided by two multiplied by 20, which in turn is three root 20 divided by 40. Noting that 20 is equal to four multiplied by five and using the laws of radicals, we can rewrite root 20 as root four multiplied by root five. Then, root 20 is equal to two root five. Our answer is therefore equal to six root five over 40. And dividing the numerator and denominator by two, we have three root five over 20.
We can therefore conclude that if 𝑓 of 𝑥 is equal to the square root of three 𝑥 minus one, then the instantaneous rate of change of 𝑓 of 𝑥 at 𝑥 is equal to seven is three root five over 20.
