Video Transcript
Find, if any, the local maximum and
local minimum values of π of π₯ is equal to five π₯ over 13 multiplied by π₯
squared plus one, together with their type.
In this question, we are given a
function π of π₯, which is a rational function. And we are asked to find the local
maximum and minimum values of this function π of π₯ together with the type of
extrema that these values are. We begin by recalling that local
extrema will always occur at the critical points of our function. And we say that π₯ is equal to π
is a critical point of the function π of π₯ if either the first derivative of π
evaluated at π is equal to zero or the derivative does not exist at this value of
π₯. This means that in order to find
the critical points of our function, weβre going to need to differentiate π of
π₯.
As already mentioned, we have a
rational function, which is the quotient of polynomials. So, we will differentiate by using
the quotient rule for differentiation. We recall the quotient rule tells
us the derivative of the quotient of two functions π’ over π£ is equal to π£ times
π’ prime minus π’ times π£ prime all divided by π£ squared. So, to apply the quotient rule to
our function π of π₯, weβll set π’ of π₯ to be the polynomial in our numerator and
weβll set π£ of π₯ to be the polynomial in our denominator. Using this, weβve rewritten π of
π₯ as π’ of π₯ divided by π£ of π₯. So, by the quotient rule, π prime
of π₯ is equal to π£ of π₯ multiplied by π’ prime of π₯ minus π’ of π₯ multiplied by
π£ prime of π₯ all divided by π£ of π₯ squared. In order to use the quotient rule,
weβll need to find expressions for π’ prime of π₯ and π£ prime of π₯.
Weβll begin with π’ prime of
π₯. This is the derivative of five π₯
with respect to π₯. As π’ of π₯ is a linear function,
its derivative is the coefficient of π₯, which in this case is five. π’ prime of π₯ is equal to
five. By distributing the parentheses on
the denominator, we can rewrite π£ of π₯ as 13π₯ squared plus 13. Differentiating this term by term,
we see that π£ prime of π₯ is equal to 26π₯, noting that differentiating a constant
gives us zero. We can now substitute our
expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯ into our
expression for π prime of π₯. This is equal to 13 multiplied by
π₯ squared plus one multiplied by five minus five π₯ multiplied by 26π₯ all divided
by 13 multiplied by π₯ squared plus one all squared.
We now want to simplify this
expression. We begin by distributing the square
over the parentheses in our denominator. This gives us a new denominator of
13 squared multiplied by π₯ squared plus one squared. Next, we can cancel a shared factor
of 13 in our numerator and denominator. We can now simplify the numerator,
giving us five π₯ squared plus five minus 10π₯ squared, which in turn is equal to
five minus five π₯ squared. π prime of π₯ is therefore equal
to five minus five π₯ squared divided by 13 multiplied by π₯ squared plus one
squared.
We can now use this expression to
find our critical points, as our critical points are where the derivative function
is equal to zero or where the derivative function does not exist. Letβs start by checking the points
where our derivative function does not exist. Since our derivative function is a
rational function, it will exist for all real values of π₯ except where the
denominator is equal to zero, that is, any values where 13 multiplied by π₯ squared
plus one all squared equals zero. However, we can show that there are
no solutions to this equation. Dividing through by 13 and square
rooting both sides, we have π₯ squared plus one equals zero. Next, we can subtract one from both
sides, such that π₯ squared is equal to negative one.
We know that this equation has no
real solutions. So, our derivative function is
defined for all real values of π₯. And we can therefore conclude that
the only critical points of our function will be where the derivative is equal to
zero. Since the derivative function is a
rational function, it will only be equal to zero if the numerator is equal to zero
and the denominator is not equal to zero. As we have already shown that our
denominator is not equal to zero for any real value of π₯, we need to solve the
equation five minus five π₯ squared is equal to zero.
We begin by adding five π₯ squared
to both sides. Dividing through by five, we have
π₯ squared is equal to one. And taking the square root of both
sides of this equation, we have π₯ is equal to positive or negative one. This means that the function will
have critical points at π₯ equals one and π₯ equals negative one. And we can calculate the values of
these by substituting π₯ equals one and π₯ equals negative one into our original
function. π of one is equal to five
multiplied by one divided by 13 multiplied by one squared plus one. And this is equal to five over
26. We can calculate π of negative one
in the same way, and this is equal to negative five over 26.
We now have the two critical values
of π of π₯, and our final step is to determine their type. One way to do this would be to use
the second derivative test. However, as our first derivative is
a rational function, in order to do this, we would need to use the quotient rule a
second time. As a result, it will be easier to
use the first derivative test. To use the first derivative test,
we need to check the derivative of our function above and below each of our critical
points. As the critical points are negative
one and one, weβll check the derivative at negative two, zero, and two. So, letβs start filling in our
table.
We know negative one and one are
critical points, and the derivative of our function π is zero at these two
points. Next, weβll find π prime evaluated
at negative two, recalling that we are only interested if this is a positive or
negative number. By evaluating this expression, we
can see the denominator is positive and the numerator is negative. This means that this is a negative
expression, and our derivative when π₯ is equal to negative two is negative. We can do the same to check our
derivative when π₯ is equal to zero. This time, our denominator is
positive and our numerator is positive. This means that our derivative is
positive when π₯ is equal to zero.
Finally, we need to check the
derivative of our function when π₯ is equal to two. If we evaluate these expressions,
we can see that our denominator is positive and our numerator is negative. This means that the derivative of
our function π of π₯ is negative when π₯ is equal to two. Letβs now consider what this means
for our critical points at negative one and one. For π₯ is equal to negative one, we
have a negative derivative around π₯ is equal to negative two. At π₯ is equal to negative one, our
derivative is zero, and our derivative goes positive around π₯ is equal to zero. This means that by the first
derivative test at negative one we have a local minimum.
We can then repeat this process for
π₯ is equal to one. Around π₯ is equal to zero, our
derivative is positive. At π₯ is equal to one, our
derivative is equal to zero. And as π₯ approaches two, our
derivative is negative. This means that by the first
derivative test at π₯ is equal to one, we have a local maximum.
We can therefore conclude that the
function π of π₯ has a local maximum equal to five over 26 and a local minimum
equal to negative five over 26. These occur when π₯ equals one and
π₯ equals negative one, respectively.