### Video Transcript

Find the intervals on which the function π of π₯ equals two π to the power of π₯ over negative three π to the power of π₯ plus four is either increasing or decreasing.

We recall that we can use calculus to identify intervals of increase or decrease of a function. We say that a function is increasing if the value of its first derivative at a given point is greater than zero. Similarly, itβs decreasing at that point if the value of the derivative there is less than zero. Of course, weβre looking to find the intervals on which our function is increasing or decreasing. In other words, the entire set of values of π₯ such that π prime of π₯ is greater than zero and the entire set of values of π₯ such that π prime of π₯ is less than zero.

So now we look at our function. We have a quotient. Itβs two π to the power of π₯ over negative three π to the power of π₯ plus four. Both the numerator and the denominator of our fraction are differentiable. And we know this because they are simple exponential functions. And exponential functions are indeed differentiable over their entire domain.

And so we can use the quotient rule to find the derivative. This says that the derivative of the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. Weβll let π’ be equal to the numerator of our fraction. Thatβs two π to the power of π₯. And therefore, π£ is equal to negative three π to the power of π₯ plus four.

We see from our rule that we need to work out dπ’ by dπ₯ and dπ£ by dπ₯. And in fact the derivative of π to the power of π₯ is simply π to the power of π₯. So dπ’ by dπ₯ is two π to the power of π₯. Similarly, the derivative of negative three π to the power of π₯ is negative three π to the power of π₯. And then the derivative of a constant is zero.

Using function notation then, we multiply π£ by dπ’ by dπ₯. Thatβs negative three π to the power of π₯ plus four times two π to the power of π₯. Then we subtract π’ times dπ£ by dπ₯. Thatβs two π to the power of π₯ times negative three π to the power of π₯. And this is all over π£ squared.

Our next job is to simplify our fraction. Weβll leave the denominator as it is for now and simply focus on the numerator. Notice we have a constant factor of two π to the power of π₯. So letβs factor this expression. Dividing negative three π to the power of π₯ plus four times two π to the power of π₯ by two π to the power of π₯ leaves us with negative three π to the power of π₯ plus four. And dividing negative two π to the power of π₯ times negative three π to the power of π₯ by two π to the power of π₯ leaves us negative negative three π to the power of π₯, which is of course simply three π to the power of π₯.

And now we notice that negative three π to the power of π₯ plus three π to the power of π₯ is zero. So our fraction becomes two π to the power of π₯ times four over negative three π to the power of π₯ plus four squared, which is eight π to the power of π₯ over negative three π to the power of π₯ plus four squared.

Now, remember, weβre looking to find the values of π₯ such that this expression is positive for an increasing function and negative for the parts of the function where itβs decreasing. And actually, we can use a bit of common sense to do so. We know that π to the power of π₯ is always positive. And that means that eight π to the power of π₯ must also be positive. It must be greater than zero.

Similarly, when we square any real number, we get a positive result. So negative three π to the power of π₯ plus four squared must be always greater than or actually equal to zero. And so since a positive divided by a positive is a positive, we can say that the first derivative will be greater than zero for all real values of π₯ except those when negative three π to the power of π₯ plus four is equal to zero.

So letβs solve the equation negative three π to the power of π₯ plus four equals zero to find the value or values of π₯ that weβre not including in our interval. We add negative three π to the power of π₯ to both sides of our equation. Then we divide through by three. So we find that π to the power of π₯ equals four-thirds.

Weβre looking to solve for π₯. And so since the natural log is the inverse of π to the power of π₯, weβre going to take the natural log of both sides of our equation. And so we find that solving this equation gives us π₯ is equal to the natural log of four-thirds. And so our function is always increasing except when π₯ is equal to this value.

We can use interval notation to represent this. Remember, if we use these curly brackets, weβre saying that π₯ can be greater than negative β and less than the natural log of four-thirds. And so our function is increasing on the open interval from negative β to the natural log of four-thirds and the open interval from the natural log of four-thirds to β.

And it is worth briefly considering what it means if we have π₯ equals the natural log of four-thirds. It doesnβt mean our first derivative is equal to zero. In fact, our first derivative at this point is equal to something divided by zero. Itβs undefined. And so we do have a critical point there. In fact, in this case, it actually represents an asymptote.