Video: Finding a Polynomial Function’s Intervals of Increase and Decrease

Determine the intervals on which the function 𝑦 = 3π‘₯Β²(9π‘₯ + 5) is increasing and where it is decreasing.

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Video Transcript

Determine the intervals on which the function 𝑦 equals three π‘₯ squared times nine π‘₯ plus five is increasing and where it is decreasing.

We begin by recalling what we actually look for to establish whether a function is increasing or decreasing. A function defined by 𝑓 of π‘₯ is increasing when its first derivative is greater than zero. And conversely it will be decreasing when the first derivative is less than zero.

Now, of course, our function is defined as 𝑦 equals three π‘₯ squared times nine π‘₯ plus five. So we’ll use Leibniz notation. And we look for d𝑦 by dπ‘₯ is greater than zero and d𝑦 by dπ‘₯ is less than zero.

So let’s find d𝑦 by dπ‘₯. We’ll differentiate our function with respect to π‘₯. Now, it’s the product of two polynomial functions. And we know that polynomial functions themselves are differentiable over their entire domain. And so the product of the functions is also differentiable over its entire domain.

And whilst it would be quite straightforward to use the product rule to find the derivative, it’s equally as simple to begin by distributing our parentheses. Three π‘₯ squared times nine π‘₯ is 27π‘₯ cubed, and three π‘₯ squared times five is 15π‘₯ squared. We’ll then differentiate term by term, recalling that the derivative of π‘Žπ‘₯ to the 𝑛th power for real constants π‘Ž and 𝑛 is 𝑛 times π‘Žπ‘₯ to the power of 𝑛 minus one. Essentially, we multiply the entire term by the exponent and then reduce that exponent by one.

So the first derivative of 27π‘₯ cubed is three times 27π‘₯ squared. And the first derivative of 15π‘₯ squared is two times 15π‘₯. And so d𝑦 by dπ‘₯ is 81π‘₯ squared plus 30π‘₯. We need to work out the values of π‘₯ such that this derivative is greater than zero and this derivative is less than zero. So how do we do this?

Well, one method is to use an inequality solver on our calculator. In the absence of a calculator though, we can consider the shape of the curve 𝑦 equals 81π‘₯ squared plus 30π‘₯. We know it’s a quadratic, and it has a positive leading coefficient, so it will be a parabola. We look for the π‘₯-intercepts of our graph by setting 81π‘₯ squared plus 30π‘₯ equal to zero then solving for π‘₯. And of course we do this by factoring the expression on the left-hand side.

We notice that 81π‘₯ squared and 30π‘₯ have a common factor of three π‘₯. So we get three π‘₯ times 27π‘₯ plus 10. And of course, for three π‘₯ times 27π‘₯ plus 10 to be equal to zero, either three π‘₯ must be equal to zero or 27π‘₯ plus 10 must be equal to zero. Solving this first equation for π‘₯, and we get π‘₯ equals zero. And then solving our second equation, we get π‘₯ equals negative 10 over 27. And so the graph of 𝑦 equals 81π‘₯ squared plus 30π‘₯ looks like this.

Remember, we’re looking to find where this is greater than zero and this is less than zero. And we see if we look carefully that this is greater than zero for values of π‘₯ less than negative 10 over 27 and greater than zero. And it’s less than zero for values of π‘₯ between negative 10 over 27 and zero. And so once we’ve sketched the curve of 𝑦 equals 81π‘₯ squared plus 30π‘₯, we’re really looking at where the derivative is greater and less than zero.

And so using interval notation, we say that our function is increasing on the open interval from negative ∞ to negative 10 over 27 and the open interval from zero to ∞. And it’s decreasing for π‘₯-values on the open interval from negative 10 over 27 to zero. And of course it’s important that we realize that these must be open intervals. We’re not interested when d𝑦 by dπ‘₯ is equal to zero, since that indicates a critical point rather than a point of increase or decrease.

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