Video Transcript
Evaluate the instantaneous rate of
change of 𝑓 of 𝑥 is equal to 𝑥 squared minus three 𝑥 plus two at 𝑥 is equal to
five.
In this question, we are given a
function 𝑓 of 𝑥 and asked to find the instantaneous rate of change when 𝑥 is
equal to five. We recall that we denote the
instantaneous rate of change of a function 𝑓 of 𝑥 at 𝑥 equals 𝑎 as 𝑓 prime
evaluated at 𝑎. And this is equal to the limit as ℎ
approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all divided by ℎ, where the limit
exists.
In this question, we want to find
the derivative when 𝑥 is equal to five. We need to calculate 𝑓 prime of
five, which is equal to the limit as ℎ approaches zero of 𝑓 of five plus ℎ minus 𝑓
of five all divided by ℎ.
We begin by finding an expression
for 𝑓 of five plus ℎ. Substituting five plus ℎ for 𝑥, we
have 𝑓 of five plus ℎ is equal to five plus ℎ squared minus three multiplied by
five plus ℎ plus two. Distributing our parentheses, this
simplifies to 25 plus 10ℎ plus ℎ squared minus 15 minus three ℎ plus two, which in
turn is equal to 12 plus seven ℎ plus ℎ squared.
Next, we’ll find the value of 𝑓 of
five. This is equal to five squared minus
three multiplied by five plus two. And 25 minus 15 plus two is equal
to 12. 𝑓 prime of five is therefore equal
to the limit as ℎ approaches zero of 12 plus seven ℎ plus ℎ squared minus 12 all
divided by ℎ. The constants on the numerator
cancel. And since ℎ approaches zero and can
therefore never be equal to zero, we can divide the numerator and denominator by
ℎ. We are left with the limit as ℎ
approaches zero of seven plus ℎ.
As we now have a polynomial in ℎ,
we can attempt to use direct substitution. Substituting ℎ is equal to zero
gives us seven plus zero, which is equal to seven. We can therefore conclude that the
instantaneous rate of change of 𝑓 of 𝑥 which is equal to 𝑥 squared minus three 𝑥
plus two at 𝑥 equals five is seven.
