Question Video: Finding the Rate of Change of a Polynomial Function at a Point | Nagwa Question Video: Finding the Rate of Change of a Polynomial Function at a Point | Nagwa

Question Video: Finding the Rate of Change of a Polynomial Function at a Point Mathematics • Second Year of Secondary School

Find the instantaneous rate of change of 𝑓(𝑥) = 5𝑥³ + 17 when 𝑥 = 3.

03:52

Video Transcript

Find the instantaneous rate of change of 𝑓 of 𝑥 equals five 𝑥 cubed plus 17 when 𝑥 equals three.

In this question, we are given a function 𝑓 of 𝑥 and are asked to find the instantaneous rate of change when 𝑥 equals three. 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 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 three. So we will set 𝑎 is equal to three in our equation for the rate of change. And this gives us the limit we need to evaluate.

To help us evaluate this limit, let′s first find an expression for 𝑓 evaluated at three plus ℎ. We substitute 𝑥 is equal to three plus ℎ into our function, giving us five multiplied by three plus ℎ cubed plus 17. We can distribute the exponent over the parentheses using our knowledge of the binomial expansion. 𝑎 plus 𝑏 cubed is equal to 𝑎 cubed plus three 𝑎 squared 𝑏 plus three 𝑎 𝑏 squared plus 𝑏 cubed. This means that cubing three plus ℎ gives us 27 plus 27ℎ plus nine ℎ squared plus ℎ cubed. We need to multiply this by five and then add 17. 𝑓 of three plus ℎ is therefore equal to 135 plus 135ℎ plus 45ℎ squared plus five ℎ cubed plus 17. And adding the constants 135 and 17 gives us 152 plus 135ℎ plus 45ℎ squared plus five ℎ cubed. Evaluating 𝑓 of three gives us five multiplied by three cubed plus 17, which is equal to 152.

We can now substitute these into our expression for 𝑓 prime of three. We have the limit as ℎ approaches zero of 152 plus 135ℎ plus 45ℎ squared plus five ℎ cubed minus 152 all divided by ℎ. 152 minus 152 is zero. We can then divide through by ℎ, giving us the limit as ℎ tends to zero of 135 plus 45ℎ plus five ℎ squared. As we now have a polynomial in ℎ, we can attempt to use direct substitution. Substituting ℎ equals zero gives us 135 plus 45 multiplied by zero plus five multiplied by zero squared. This means that the second and third terms cancel, and we are just left with 135. And we can therefore conclude that the instantaneous rate of change of 𝑓 of 𝑥 equals five 𝑥 cubed plus 17 when 𝑥 equals three is 135.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy