When working with function composition, we are essentially plugging one function in as the input of another function. We then simplify and give our answer. When we see (f ○ g)(x) , f(g(x)), or f[g(x)], we are being asked to plug the function g(x) in for x in the function f(x).

Test Objectives
• Demonstrate a general understanding of function notation
• Demonstrate the ability to find the composition of two or more functions
• Demonstrate the ability to evaluate a composite function
• Demonstrate the ability to find the domain for a composite Function
Function Composition Practice Test:

#1:

Instructions: find f(g(x)), f(g(2)), and f(g(x + 1)).

$$a)\hspace{.2em}$$ $$f(x)=2x - 1$$ $$g(x)=5x + 1$$

Instructions: find g(f(x)), g(f(-3)), and g(f(a - 1)).

$$b)\hspace{.2em}$$ $$f(x)=2x^2 - 5$$ $$g(x)=\frac{x}{7}$$

#2:

Instructions: find f(g(x)) and f(g(x2)).

$$a)\hspace{.2em}$$ $$f(x)=x^2 - 7x - 3$$ $$g(x)=\frac{2x}{3}$$

$$b)\hspace{.2em}$$ $$f(x)=x^3 - 2x$$ $$g(x)=x - 2$$

#3:

Instructions: find f(g(x)), g(f(x)), and state the domain for each.

$$a)\hspace{.2em}$$ $$f(x)=x^2$$ $$g(x)=\sqrt{x - 1}$$

$$b)\hspace{.2em}$$ $$f(x)=\frac{1}{x + 2}$$ $$g(x)=\frac{3}{x - 1}$$

#4:

Instructions: find f(g(h(x))) and state the domain.

$$a)\hspace{.2em}$$ $$f(x)=x - 1$$ $$g(x)=\sqrt{x}$$ $$h(x)=x - 1$$

$$b)\hspace{.2em}$$ $$f(x)=\frac{1}{x}$$ $$g(x)=\sqrt{x + 2}$$ $$h(x)=\frac{x}{x - 1}$$

#5:

$$a)\hspace{.2em}$$ $$f(g(x))=(x - 3)^2 - 5(x - 3) + 1$$

$$b)\hspace{.2em}$$ $$f(g(x))=\frac{1}{x^2 - 3x + 2}$$

#1:

Solutions:

$$a)\hspace{.2em}$$ $$f(g(x))=10x + 1$$ $$f(g(2))=21$$ $$f(g(x + 1))=10x + 11$$

$$b)\hspace{.2em}$$ $$g(f(x))=\frac{2x^2 - 5}{7}$$ $$g(f(-3))=\frac{13}{7}$$ $$g(f(a - 1))=\frac{2a^2 - 4a - 3}{7}$$

#2:

Solutions:

$$a)\hspace{.2em}$$ $$f(g(x))=\frac{4x^2 - 42x - 27}{9}$$ $$f(g(x^2))=\frac{4x^4 - 42x^2 - 27}{9}$$

$$b)\hspace{.2em}$$ $$f(g(x))=x^3 - 6x^2 + 10x - 4$$ $$f(g(x^2))=x^6 - 6x^4 + 10x^2 - 4$$

#3:

Solutions:

$$a)\hspace{.2em}$$ $$f(g(x))=x - 1$$ $$\text{Domain:} \, \{x | x ≥ 1 \}$$ $$g(f(x)) = \sqrt{x^2 - 1}$$ $$\text{Domain:} \, \{x | x ≤ -1, x ≥ 1\}$$

$$b)\hspace{.2em}$$ $$f(g(x)) = \frac{x + 1}{2x + 1}$$ $$\text{Domain:} \, \left\{x \middle| x ≠ -\frac{1}{2}, 1\right\}$$ $$g(f(x))=-\frac{3(x + 2)}{x + 1}$$ $$\text{Domain:} \, \{x | x ≠ -2, -1\}$$

#4:

Solutions:

$$a)\hspace{.2em}$$ $$f(g(h(x))) = \sqrt{x - 1} - 1$$ $$\text{Domain:} \, \{x | x ≥ 1\}$$

$$b)\hspace{.2em}$$ $$f(g(h(x)) = \frac{1}{\large{\sqrt{\frac{3x-2}{x-1}}}}$$ $$\text{Domain:} \, \left\{x \middle| x < \frac{2}{3}, x > 1\right\}$$ Note: If you want to rationalize the denominator, you need to be very careful that you don't create a function that is different than the original when x < 2/3.
Desmos Link for More Detail Notice the function graph below does not match the original function graph given above. This is coming from the fact that when x < 2/3, we have a negative y-value, whereas in the original we had a positive y-value.
Desmos Link for More Detail You can write this as a piecewise-defined function if you want. I would note that this is not necessary as not rationalizing the denominator for this problem is the normal way to answer. Basically, the idea is to flip the sign back to what it should be when x < 2/3. $$f(g(h(x))) = \begin{cases} \frac{\sqrt{(3x - 2)(x - 1)}}{3x - 2} & \text{if} \, x > 1 \\ -\frac{\sqrt{(3x - 2)(x - 1)}}{3x - 2} & \text{if} \, x < \frac{2}{3} \end{cases}$$ This will match the original function and its domain, allowing you to rationalize the denominator. You can see the final graph is a match of the first one.
Additionally, you could write the denominator using absolute value bars. $$f(g(h(x))) = \frac{\sqrt{(3x - 2)(x - 1)}}{|3x - 2|}$$ No matter which form you choose, you will always keep the same domain from before you did any work to rationalize the denominator. $$\text{Domain:} \, \left\{x \middle| x < \frac{2}{3}, x > 1\right\}$$

#5:

Solutions:

$$a)\hspace{.2em}$$ Your answer may vary. $$f(x)=x^2 - 5x + 1$$ $$g(x)=x - 3$$

$$b)\hspace{.2em}$$ Your answer may vary. $$f(x) = \frac{1}{x}$$ $$g(x) = x^2 - 3x + 2$$