Video Transcript
Determine the intervals on which the function 𝑓 of 𝑥 equals 𝑥 plus three times the absolute value of 𝑥 plus three is increasing and decreasing.
Remember, we can calculate intervals of increase and decrease for a function by thinking about its derivative. Specifically, if we can find the derivative of that function over some interval and the value of that derivative is positive — it’s greater than zero — then the function itself must be increasing on that interval. Similarly, if the derivative is negative, if it’s less than zero over some interval, then the function must be decreasing over that interval. And of course we’re going to need to be a little bit careful here. This function is the product of a linear function and the modulus of a linear function. So how do we calculate its derivative?
Well, let’s begin by reminding ourself what the modulus function does. This function specifically takes values of 𝑥, then adds three to that value, then makes sure that the output is positive. For instance, suppose 𝑥 is equal to negative five. Negative five plus three is negative two. And then the modulus or the absolute value of that is positive two.
Now, it follows for values of 𝑥 less than negative three, the value of 𝑥 plus three is negative. So when 𝑥 is less than negative three, the output of the function, the absolute value of 𝑥 plus three, is negative of 𝑥 plus three. But when 𝑥 is greater than or equal to negative three, the value of 𝑥 plus three is positive. So when 𝑥 is greater than or equal to negative three, the absolute value simply outputs 𝑥 plus three. This means we’re allowed to think of the absolute value function as a piecewise-defined function. And in turn it also means that we can rewrite 𝑓 of 𝑥 using piecewise notation as shown.
When 𝑥 is less than negative three, the function is the product of 𝑥 plus three and negative 𝑥 plus three. And when 𝑥 is greater than or equal to negative three, we get 𝑥 plus three times 𝑥 plus three. And we can of course simplify each part of this function as shown. Then, to find the derivative of 𝑓 of 𝑥, we’re going to differentiate each part of the function.
Now, 𝑥 plus three squared and negative 𝑥 plus three squared are polynomials. This means they are differentiable over their entire domain, which is of course the set of real numbers. So we can use the general power rule to differentiate negative 𝑥 plus three squared over the interval 𝑥 is less than negative three and to differentiate 𝑥 plus three squared over the interval 𝑥 is greater than or equal to negative three.
The generalized power rule for differentiation is a special version of the chain rule. It says that we can differentiate the 𝑛th power of 𝑓 of 𝑥 by multiplying 𝑛 times 𝑓 of 𝑥 to the power of 𝑛 minus one by the derivative of the inner function, so 𝑓 prime of 𝑥. In other words, the derivative of negative 𝑥 plus three squared is two times negative 𝑥 plus three to the first power times the derivative of 𝑥 plus three, which is one. We can rewrite that of course as negative two times 𝑥 plus three to the first power times one.
In a similar way, the derivative of 𝑥 plus three squared is two times 𝑥 plus three to the first power times the derivative of 𝑥 plus three, which is one. And then we can simplify each part of this expression as shown. Now, it’s worth noting that this process works because the function itself is continuous at 𝑥 equals negative three. But also both the left and right derivatives exist and in fact are equal to zero at 𝑥 equals negative three.
Now let’s consider the sign of the derivative over each interval. When 𝑥 is less than negative three, we get a negative times a negative, which is of course a positive. Then, when 𝑥 is greater than or equal to negative three, we get a positive multiplied by a nonnegative, which gives a nonnegative. It’s greater than or equal to zero. We can therefore say that the derivative is positive for all real values of 𝑥 except when 𝑥 is equal to negative three. But at that point, it’s equal to zero. We’re allowed to include this value in the interval of increase if we choose. So we can deduce that 𝑓 prime of 𝑥 is greater than or equal to zero for all real numbers. And this, in turn, means that 𝑓 of 𝑥 is increasing over the set of real numbers.
