Video: Finding Intervals of Increase of a Function

Consider a function whose first derivative is 𝑓′(π‘₯) = ((π‘₯ + 11)(π‘₯ βˆ’ 7))/(π‘₯ βˆ’ 3). On which interval(s) is 𝑓 increasing?

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Video Transcript

Consider a function whose first derivative is 𝑓 prime of π‘₯ equals π‘₯ plus 11 times π‘₯ minus seven over π‘₯ minus three. On which interval or intervals is 𝑓 increasing?

Let’s consider what it actually means for a function to be increasing. A function is increasing when its rate of change is positive. In terms of its derivatives, we say that a function is increasing when its first derivative is greater than zero.

Now, we’ve actually been given the first derivative of our function 𝑓. We’re told that 𝑓 prime of π‘₯ is π‘₯ plus 11 times π‘₯ minus seven over π‘₯ minus three. Our job then is to find the value or values of π‘₯ such that this is greater than zero.

Now, we’ll need to be really careful. This is a rational function. So for the quotient of two functions to be greater than zero, both functions themselves must be greater than zero. Or both functions must be less than zero since we know a negative divided by a negative is a positive.

Let’s look at the first scenario. We’re interested in the values of π‘₯ such that π‘₯ plus 11 times π‘₯ minus seven is greater than zero and π‘₯ minus three is greater than zero. In other words, we’re looking for the intersection of the solutions. Now, one way we can find the values of π‘₯ such that π‘₯ plus 11 times π‘₯ minus seven is greater than zero is to first consider where this is equal to zero. Well, for the product of two values to be zero, either one of those values must themselves be zero. So we can say that π‘₯ plus 11 times π‘₯ minus seven will be equal to zero when π‘₯ plus 11 is equal to zero or π‘₯ minus seven is equal to zero. Solving each equation, and we see that this is when π‘₯ is equal to negative 11 and π‘₯ is equal to seven.

We know that the graph of 𝑦 equals π‘₯ plus 11 times π‘₯ minus seven is a quadratic curve. And we now know it passes through the π‘₯-axis at negative 11 and seven. The graph is greater than zero. And therefore, π‘₯ plus 11 times π‘₯ minus seven is greater than zero in these two locations. So, in other words, when π‘₯ is less than negative 11 or greater than seven.

But if we go back to our other inequality β€” π‘₯ minus three is greater than zero β€” and we add three to both sides, we see that π‘₯ must be greater than three. Remember, we’re looking for the intersection of these solutions. Well, π‘₯ cannot be greater than three and less than negative 11. It also can’t purely be greater than three and greater than seven. So the intersection is in fact π‘₯ is greater than seven.

In interval notation then, we say that 𝑓 is increasing on the open interval from seven to positive ∞. But remember, we also said we needed to look for the values of π‘₯ such that both the numerator and denominator of our fraction are negative. Well, this time, it’s the other part of the graph we’re interested in that will tell us the values of π‘₯ such that π‘₯ plus 11 times π‘₯ minus seven is less than zero. In this case, it’s values of π‘₯ greater than negative 11 and less than seven.

But if we go to our second inequality, we see that π‘₯ is less than three. So what’s the intersection? What’s the overlap of our solutions here? Well, the intersection the values of π‘₯ which satisfy both sets of inequalities is π‘₯ is greater than negative 11 and less than three. In interval notation, that’s the open interval from negative 11 to three.

And so the function 𝑓 is increasing on the open interval from seven to positive ∞ and negative 11 to three.

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