# Video: Finding the Intervals Where Functions Involving Logarithmic Functions Increase and Decrease

Given that 𝑓(𝑥) = 5𝑥² − 3𝑥 − ln 𝑥, find the intervals on which 𝑓 is increasing or decreasing.

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### Video Transcript

Given that 𝑓 of 𝑥 equals five 𝑥 squared minus three 𝑥 minus the natural log of 𝑥, find the intervals on which 𝑓 is increasing or decreasing.

First, we recall that a function is increasing when its first derivative 𝑓 prime of 𝑥 is greater than zero, and a function is decreasing when its first derivative 𝑓 prime of 𝑥 is less than zero. We’re therefore going to need an expression for the first derivative of this function. We can differentiate term-by-term. The derivative of five 𝑥 squared is five multiplied by two 𝑥; that’s 10𝑥. The derivative of negative three 𝑥 is negative three. And the derivative of negative the natural log of 𝑥 is negative one over 𝑥. So, we have our first derivative. 𝑓 prime of 𝑥 equals 10𝑥 minus three minus one over 𝑥.

By our definition of an increasing function, first of all, 𝑓 will be increasing when its first derivative 10𝑥 minus three minus one over 𝑥 is greater than zero. And therefore, we have an inequality in 𝑥 that we need to solve. Now, we know that there is an 𝑥 in the denominator of this fraction here, so the step we’d like to take first of all is to multiply by 𝑥 in order to eliminate this fraction. But we need to be a little bit careful because we have an inequality and no guarantee that 𝑥 is positive. If we multiply by a negative 𝑥-value, then we would need to reverse the direction of our inequality.

However, if we look back at our original function, we see that it includes this term the natural log of 𝑥. And the natural logarithm of 𝑥 is undefined for 𝑥-values less than or equal to zero. This means that the domain of our function 𝑓 of 𝑥 is 𝑥 greater than zero. We’re only working with positive values of 𝑥. And therefore, we can multiply our inequality by 𝑥 without worrying about needing to change the direction of the inequality sign.

Multiplying by 𝑥 gives 10𝑥 squared minus three 𝑥 minus one is greater than zero. And we see that we have a quadratic inequality which we need to solve. There are a number of different methods that we can use, but almost certainly we need to factorise to begin with. By following the formal method of factoring by grouping, all with a bit of trial and error, we see that this quadratic factors as five 𝑥 plus one multiplied by two 𝑥 minus one.

We then need to find the critical values for this quadratic, which we do by setting each of our two brackets equal to zero, not greater than zero. We then solve each linear equation to give 𝑥 equals negative one-fifth and 𝑥 equals one-half. So, these are the two critical values for this quadratic. Now, there are two ways that we can proceed from here. One is to use a table of values to check the sign of our quadratic either side and in-between our critical values. The other is to sketch a graph. And that’s the one that I’m going to choose to demonstrate.

We know that we have a quadratic with a positive leading coefficient. So, its graph will be a parabola. And we know the critical values, which are the values at which the graph crosses the 𝑥-axis, are negative one-fifth and one-half. So, the graph looks like this. Remember, this is the graph of our first derivative, 10𝑥 minus three minus one over 𝑥. We said that our function 𝑓 will be increasing when its first derivative 𝑓 prime of 𝑥 is greater than zero. That is when the graph of its derivative 𝑓 prime of 𝑥 is above the 𝑥-axis.

This will correspond to two sections of our graph, the section where 𝑥-values are less than negative one-fifth and the part where 𝑥-values are greater than one-half. But remember, we said that the domain of our function 𝑓 of 𝑥 was just 𝑥 greater than zero. And therefore, we can actually ignore one-half of our graph completely. We can say then that our function 𝑓 is increasing on the open interval one-half, infinity. That’s all 𝑥-values greater than one-half.

To see where our function is decreasing, we’re looking at where the first derivative 𝑓 prime of 𝑥 is less than zero, which means we’re looking at where its graph is below the 𝑥-axis. Now, on our original graph, this would’ve been everywhere between the two critical values. But as we’ve reduced the graph to be only 𝑥-values greater than zero, this is for all 𝑥-values greater than zero but less than one-half. We therefore say that 𝑓 is decreasing on the open interval zero, one-half.

So, we’ve completed the problem. We had to differentiate the function 𝑓 of 𝑥 to find its first derivative 𝑓 prime of 𝑥, and then use our knowledge of quadratic inequalities to find where 𝑓 prime of 𝑥 was greater than zero and where 𝑓 prime of 𝑥 was less than zero.