# Question Video: Finding the Intervals of Increasing and Decreasing of a Polynomial Function Mathematics • Higher Education

Determine the intervals over which the function π(π₯) = 11π₯Β³ β 8π₯Β² is increasing and over which it is decreasing.

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### Video Transcript

Determine the intervals over which the function π of π₯ is equal to 11π₯ cubed minus eight π₯ squared is increasing and over which it is decreasing.

The question gives us a polynomial function π of π₯. It wants us to find the intervals over which this function is increasing and the intervals over which this function is decreasing. And we recall for a differentiable function π, we say itβs increasing when its derivative is greater than zero. Intuitively, this is because its slope will be greater than zero, so its slope is pointing upwards. So, itβs getting larger. And we also know that for a differentiable function π, we say itβs decreasing when its derivative is less than zero. Since our function π of π₯ is a polynomial, itβs differentiable for all real numbers. So, we want to find the intervals which contain those values of π₯ such that the derivative of π is greater than zero and the derivative of π is less than zero.

Letβs start by finding our derivative function π prime of π₯. Since weβre differentiating a polynomial, we can do this by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. Differentiating our polynomial term by term using this rule, we get three times 11 times π₯ to the power of three minus one minus two times eight times π₯ to the power of two minus one, which simplifies to give us 33π₯ squared minus 16π₯. So, our derivative function π prime of π₯ is a quadratic. And we want to find the values of π₯ where this quadratic is greater than zero and the values of π₯ where this quadratic is less than zero. Thereβs a few different methods we could use. Weβre going to do this by sketching a graph of π prime of π₯.

So, we want to sketch a graph of the quadratic π¦ is equal to 33π₯ squared minus 16π₯. We see the leading term of our polynomial is 33π₯ squared. Since 33 is positive, our sketch should have a similar shape to the parabola π¦ is equal to π₯ squared. Next, we can find the values of our π₯-intercepts by factoring this equation. We see that both terms share a factor of π₯. Taking out the shared factor of π₯, we get π₯ multiplied by 33π₯ minus 16. And we can find the π₯-intercepts of our sketch by solving each factor equal to zero. This gives us π₯-intercepts of π₯ is equal to zero and π₯ is equal to 16 divided by 33.

We can now sketch a graph of our derivative function, π¦ is equal to 33π₯ squared minus 16π₯. We mark our π₯-intercepts, zero and 16 divided by 33. And we know that our sketch of π¦ is equal to π prime of π₯ should have a similar shape to the parabola π¦ is equal to π₯ squared. We want to use this sketch to find the values of π₯ where π prime of π₯ is bigger than zero and where π prime of π₯ is less than zero. This is the same as finding the values of π₯ where our curve lies above and below the π₯-axis.

We can see if π₯ is less than zero, our curve lies above the π₯-axis. And if π₯ is greater than 16 over 33, our curve also lies above the π₯-axis. So, we have when π₯ is less than zero, π prime of π₯ is greater than zero. And when π₯ is greater than 16 over 33, π prime of π₯ is also greater than zero. Saying π₯ is less than zero is the same as saying that π₯ is in the open interval from negative β to zero. And saying π₯ is greater than 16 over 33 is the same as saying π₯ is in the open interval from 16 over 33 to β. So, we found the intervals where our function π of π₯ is increasing. We can do the same to find the intervals where π of π₯ is decreasing.

From our sketch, we can see that π prime of π₯ is below the π₯-axis when π₯ is between zero and 16 over 33. So, when π₯ is greater than zero and π₯ is less than 16 over 33, we have that π prime of π₯ is less than zero. This tells us that our function π of π₯ is decreasing for all values of π₯ on the open interval from zero to 16 over 33. Therefore, weβve shown that the function π of π₯ is equal to 11π₯ cubed minus eight π₯ squared is decreasing over the open interval from zero to 16 over 33 and increasing over the open intervals from negative β to zero and from 16 over 33 to β.