Course Notes for Calculus I

For the composition \(f \circ g\text{,}\) \(f\) is the outside function and \(g\) is the inside function. I use the function \(g\) to replace the variable in the function \(f\text{.}\)

In the resulting function, I need to exclude \(x = 0\) to avoid division by zero. The inside function \(g\) has no restrictions, so there are no other domain concerns and the domain is all real numbers except 0.

For the composition \(g \circ f\text{,}\) \(g\) is the outside function and \(f\) is the inside function. I use the function \(f\) to replace the variable in the function \(g\text{.}\)

In the resulting function, I need to exclude \(x = -4\) to avoid division by zero. The inside function \(f\) has the same restriction, \(x \neq -4\text{,}\) so there are no other domain concerns and the domain is all real numbers except \(-4\text{.}\)

For the composition \(f \circ g\text{,}\) \(f\) is the outside function and \(g\) is the inside function. I use the function \(g\) to replace the variable in the function \(f\text{.}\)

The resulting function has no obvious domain restrictons. However, since I have to do the inside function \(g\) first, I need to respect the domain restrictions of \(g\text{.}\) The argument in the square root needs to be positive, I need \(x^3-5 \geq 0\text{.}\) This inequality simplifies into \(x^3 \geq 5\) and further to \(x \geq \sqrt[3]{5}\text{.}\) Therefore, the domain of the composition is all real numbers \(x \geq \sqrt[3]{5}\text{.}\)

For the composition \(g \circ f\text{,}\) \(g\) is the outside function and \(f\) is the inside function. I use the function \(f\) to replace the variable in the function \(g\text{.}\)

I’ve simplified this one step using the laws of exponents. Even though this is still a complicated expression, there aren’t any more reasonable simplifications. The domain restrictions are also complicated here. The inside function \(f\) has no restrictions, so I only have to worry about the final form; unfortunately, that final form isn’t easy to pull apart. The square root is the only restriction, so the term inside the square root must be non-negative. I will work with that inequality a bit. (I’m going to apply a logarithm to get ride of the exponential; this respects the inequality, since the logarithm is an increasing function. Any increasing function preserves inequalities and an decreasing function reverses them.)

After the manipulation of this inequality, I get an understandable domain restriction. This composition is defined for all real numbers \(x \geq \sqrt{\frac{1}{3} \ln 5 + 1}\text{.}\)

I want to write \(f = g \circ h\) for some functions \(g\) and \(h\text{.}\) I need to break this into successive processes. The first observation is that the fourth root is the last operations of the function: first I square the variable, then subtract 4, then apply the fourth root. Therefore, the fourth root is a reasonable choice of the outside function, with everything inside the root as the inside function. That gives a decomposition.

Note that this isn’t the only possible answer. Often there are many ways to break apart a function as the composition of two (or more) pieces. Here is another possibility.

I want to write \(f = g \circ h\) for some functions \(g\) and \(h\text{.}\) I need to break this into successive processes. If I want the simplest outside function, I can observe that the sine function is the last step, applied to everything else in the function. I could choose the sine function as the outside function, leaving everything else as the inside.

There are other possible solutions. Here is one where I would like the inside function to be slightly simpler.

The function is increasing on this domain (since it is the composition of two increasing functions), so it is invertible. I follow the steps of the process: set \(y = f(x)\text{,}\) solve for \(y\text{,}\) then switch variables to write the inverse as a function of the conventional independent variable \(x\text{.}\)

Note that I use the exponential to remove the logarithm: these are inverse operations, so one removes the other.

When \(x > 2 \text{,}\) the logarithm \(\ln (x-1)\) is positive and growing, so its reciprocal is positive and decreasing. The function is invertible on this domain. I follow the steps of the process: set \(y = f(x)\text{,}\) solve for \(y\text{,}\) then switch variables to write the inverse as a function of the conventional independent variables \(x\text{.}\) (In the step from the first to the second line, we simply take the reciprocal of each side of the equation; when the numerators are not zero, this is a valid operation).