Determine the intervals on which
the function 𝑓 of 𝑥 equals seven 𝑥 over 𝑥 squared plus nine is increasing and
where it is decreasing.
We recall, first of all, that
whether a function is increasing or decreasing can be determined by considering its
derivative. A function will be increasing when
its first derivative is positive, and it will be decreasing when its first
derivative is negative. We therefore need to find an
expression for 𝑓 prime of 𝑥. We note, first of all, that 𝑓 is a
quotient. So, in order to find this
derivative, we’re going to need to apply the quotient rule.
The quotient rule tells us that for
two differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their
quotient, 𝑢 over 𝑣, is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all
over 𝑣 squared. We therefore let 𝑢 equal the
numerator of our quotient, that’s seven 𝑥, and 𝑣 equal the denominator, that’s 𝑥
squared plus nine. d𝑢 by d𝑥 and d𝑣 by d𝑥 can each be found using the power rule
of differentiation. d𝑢 by d𝑥 is seven, and d𝑣 by d𝑥 is two 𝑥.
Substituting into the formula for
the quotient rule, we have then that 𝑓 prime of 𝑥 is equal to 𝑥 squared plus nine
multiplied by seven minus seven 𝑥 multiplied by two 𝑥 all over 𝑥 squared plus
nine all squared. Distributing the parentheses in the
numerator gives seven 𝑥 squared plus 63 minus 14𝑥 squared all over 𝑥 squared plus
nine all squared. Which then simplifies to 63 minus
seven 𝑥 squared over 𝑥 squared plus nine all squared. And so, we have our expression for
the first derivative.
Our function 𝑓 will be increasing
when its first derivative is greater than zero. So, we have an inequality in 𝑥
that we need to solve. Now, we can actually simplify this
somewhat. Note that the denominator of this
fraction is something squared, 𝑥 squared plus nine all squared. And therefore, the denominator
itself will always be greater than zero. In order for the whole fraction to
be greater than zero then, we only need to ensure that its numerator is greater than
zero because a positive divided by a positive will give something which is
The inequality, therefore,
simplifies to 63 minus seven 𝑥 squared is greater than zero. We can divide through by seven and
then add 𝑥 squared to each side to give nine is greater than 𝑥 squared. Or written the other way around, 𝑥
squared is less than nine. So, we have a relatively
straightforward quadratic inequality to solve. If 𝑥 squared must be less than
nine, that’s strictly less than nine, then we can have any 𝑥-value between negative
three and three, although not including these values themselves. The solution to this quadratic
inequality then is negative three is less than 𝑥 is less than three, or the open
interval negative three to three.
So, we found the only interval on
which the function 𝑓 of 𝑥 is increasing. To determine where the function is
decreasing, we require its first derivative to be less than zero, which in turn
leads to 63 minus seven 𝑥 squared is less than zero. We therefore reverse the direction
of all the inequality signs in our previous working out leading to 𝑥 squared is
greater than nine. This is only the case for 𝑥-values
strictly less than negative three and 𝑥-values strictly greater than positive
three. So, we find that there were two
intervals on which our function is decreasing. The open intervals negative
infinity to negative three and three, infinity.
So, by applying the quotient rule
to find the first derivative of our function 𝑓 of 𝑥, and then solving a relatively
straightforward quadratic inequality. We find that the function 𝑓 of 𝑥
is increasing on the open interval negative three to three and decreasing on the
open intervals negative infinity to negative three and three, infinity.