Video Transcript
Find the intervals on which the function 𝑓 of 𝑥 is equal to five 𝑥 times the square root of negative five 𝑥 plus three is increasing and decreasing.
Let’s begin by recalling what it means for a function to be increasing or decreasing. If we think about this visually, a function is increasing if the graph of that function is sloping upwards. The slope is represented by the derivative 𝑓 prime of 𝑥. And so we need the derivative 𝑓 prime of 𝑥 to be greater than zero for the function to be increasing.
The converse is true for a function that is decreasing. Its first derivative 𝑓 prime of 𝑥 will be less than zero. Note that we’re not interested at the point where the first derivative is equal to zero as that represents a stationary returning point. And so it should be quite clear that we’re going to need to find 𝑓 prime of 𝑥. We’re going to need to differentiate our function.
Our function is itself the product of two differentiable functions. And so we’re going to use the product rule. This says that the first derivative of 𝑢 times 𝑣 with respect to 𝑥 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. And so let’s define 𝑢 as being equal to five 𝑥 and 𝑣 as being equal to the square root of negative five 𝑥 plus three. That’s negative five 𝑥 plus three to the power of one-half. The product rule shows us that we’re going to need to differentiate each of these with respect to 𝑥.
Well, the first derivative of five 𝑥 is simply five. So d𝑢 by d𝑥 is equal to five. We use a special version of the chain rule to differentiate negative five 𝑥 plus three to the power of one-half. This is called the general power rule. What we do is we multiply the entire term by the exponent and then reduce the exponent by one. We then multiply all of that by the derivative of the inner function. And the derivative of negative five 𝑥 plus three is negative five. So d𝑣 by d𝑥 is negative five over two times negative five 𝑥 plus three.
And since a negative exponent gives us a reciprocal, we can write this as negative five over two times the square root of negative five 𝑥 plus three. Then 𝑓 prime of 𝑥 is 𝑢 times d𝑣 by d𝑥. So that’s five 𝑥 times negative five over two times the square root of negative five 𝑥 plus three plus 𝑣 times d𝑢 by d𝑥. That’s the square root of negative five 𝑥 plus three times five. This simplifies to negative 25𝑥 over two times the square root of negative five 𝑥 plus three plus five times the square root of negative five 𝑥 plus three.
In order to work out where this derivative is greater than or less than zero, we’re going to need to simplify this expression. And so we’ll treat our second term as a fraction and create a common denominator. We’re going to need to multiply the numerator and denominator of our second fraction by two times the square root of negative five 𝑥 plus three. The square root of negative five 𝑥 plus three times itself is simply negative five 𝑥 plus three. So we get 10 times negative five 𝑥 plus three over two times the square root of negative five 𝑥 plus three. And we can now add these numerators.
Doing so at the same time as distributing the parentheses, and we find that our fraction is negative 25𝑥 minus 50𝑥 plus 30 over two times the square root of negative five 𝑥 plus three. And that simplifies to negative 75𝑥 plus 30 over two times the square root of negative five 𝑥 plus three. Now we’re ready to find the intervals on which our function is increasing and decreasing. Before we do though, we need to add a criteria on 𝑥 to ensure that we’re not working with a negative square root. Remember, the square root of a negative number is not going to give us a real value. So we need the expression inside the square root, negative five 𝑥 plus three, to be greater than zero.
If we subtract three from both sides, we get negative five 𝑥 is greater than negative three. And then we divide through by negative five, remembering to reverse the inequality symbol when we do so. So we find that no matter what, our value of 𝑥 must be less than three over five for our derivative to be defined. And so let’s take the case when our function is increasing. We need its first derivative to be greater than zero. The first derivative is a fraction though. We know that the square root of negative five 𝑥 plus three for values of 𝑥 less than three-fifths will itself always be positive. So two times this number will also be positive. So the fraction will be positive if the numerator is positive. And this is because a positive divided by a positive is a positive.
So our function will be increasing when negative 75𝑥 plus 30 is greater than zero. We subtract 30 from both sides, so negative 75𝑥 is greater than negative 30. And then we divide through by negative 75, once again reversing that inequality symbol because we’re dividing by a negative number. And we get 𝑥 is less than 30 over 75, which simplifies to two-fifths. And so the function is increasing for values of 𝑥 less than two-fifths. Two-fifths is less than three-fifths. So we don’t worry about our earlier criteria. In set notation, we can say therefore that the function is increasing in the open interval negative ∞ to two-fifths.
Now of course, for the function to be decreasing, its first derivative must be less than zero. Since the denominator is always positive, the numerator itself must be negative for this to be the case. In other words, negative 75𝑥 plus 30 is less than zero. This time, solving this inequality for 𝑥 and we find 𝑥 is greater than two-fifths. Remember of course though 𝑥 must be less than three-fifths for our derivative to be defined and in fact for our original function to be defined. So the function is decreasing over the open interval two-fifths to three-fifths.
The intervals on which the function is increasing and decreasing are, respectively, the open interval negative ∞ to two-fifths and the open interval two-fifths to three-fifths.
