# Video: Finding the Intervals of Increasing and Decreasing of a Rational Function

Determine the intervals on which the function π(π₯) = 7π₯/(π₯Β² + 9) is increasing and where it is decreasing.

04:12

### Video Transcript

Determine the intervals on which the function π of π₯ equals seven π₯ over π₯ 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 π prime of π₯. We note, first of all, that π is a quotient. So, in order to find this derivative, weβre going to need to apply the quotient rule.

The quotient rule tells us that for two differentiable functions π’ and π£, the derivative with respect to π₯ of their quotient, π’ over π£, is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. We therefore let π’ equal the numerator of our quotient, thatβs seven π₯, and π£ equal the denominator, thatβs π₯ squared plus nine. dπ’ by dπ₯ and dπ£ by dπ₯ can each be found using the power rule of differentiation. dπ’ by dπ₯ is seven, and dπ£ by dπ₯ is two π₯.

Substituting into the formula for the quotient rule, we have then that π prime of π₯ is equal to π₯ squared plus nine multiplied by seven minus seven π₯ multiplied by two π₯ all over π₯ squared plus nine all squared. Distributing the parentheses in the numerator gives seven π₯ squared plus 63 minus 14π₯ squared all over π₯ squared plus nine all squared. Which then simplifies to 63 minus seven π₯ squared over π₯ squared plus nine all squared. And so, we have our expression for the first derivative.

Our function π will be increasing when its first derivative is greater than zero. So, we have an inequality in π₯ that we need to solve. Now, we can actually simplify this somewhat. Note that the denominator of this fraction is something squared, π₯ 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 positive.

The inequality, therefore, simplifies to 63 minus seven π₯ squared is greater than zero. We can divide through by seven and then add π₯ squared to each side to give nine is greater than π₯ squared. Or written the other way around, π₯ squared is less than nine. So, we have a relatively straightforward quadratic inequality to solve. If π₯ squared must be less than nine, thatβs strictly less than nine, then we can have any π₯-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 π₯ is less than three, or the open interval negative three to three.

So, we found the only interval on which the function π of π₯ 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 π₯ squared is less than zero. We therefore reverse the direction of all the inequality signs in our previous working out leading to π₯ squared is greater than nine. This is only the case for π₯-values strictly less than negative three and π₯-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 π of π₯, and then solving a relatively straightforward quadratic inequality. We find that the function π of π₯ is increasing on the open interval negative three to three and decreasing on the open intervals negative infinity to negative three and three, infinity.