Composite Functions

Let $f_1 : (0, 2\pi) \to R$, $f_1(x) = \sin(x)$, and $g(x) = \sin(2x)$. Find all values of $x$ in the interval $(0, 2\pi)$ for which the composition $f_1 \circ g$ is defined.

1. 1

   The composite $f_1 \circ g$ requires the inner output $g(x) = \sin(2x)$ to land in the domain $(0, 2\pi)$ of $f_1$. Since $\sin(2x) \in [-1, 1]$ always, the only binding condition is $\sin(2x) > 0$.

   $$\sin(2x) \in (0, 2\pi) \cap [-1, 1] \implies \sin(2x) > 0$$
2. 2

   Solve $\sin(2x) > 0$ for $x \in (0, 2\pi)$. Sine is positive when its argument lies in $(0, \pi)$ or $(2\pi, 3\pi)$, so $2x \in (0, \pi) \cup (2\pi, 3\pi)$.

   $$2x \in (0, \pi) \cup (2\pi, 3\pi)$$
3. 3

   Divide through by $2$ to recover the $x$ intervals. The examiner report notes many students knew $\sin(2x) > 0$ was required but could not produce the correct intervals.

   $$x \in \left(0, \tfrac{\pi}{2}\right) \cup \left(\pi, \tfrac{3\pi}{2}\right)$$

VCAA 2024 Mathematical Methods Exam 2, Section B Q5d

Let $f(x) = \log_e x$, where $x > 0$, and $g(x) = \sqrt{1 - x}$, where $x < 1$. Find the domain of the derivative of $(f \circ g)(x)$.

1. 1

   Build the composite. $(f \circ g)(x) = \log_e \sqrt{1 - x}$ requires the inner output to be a legal input for the log, so $\sqrt{1 - x} > 0$, which forces $x < 1$.

   $$(f \circ g)(x) = \log_e \sqrt{1 - x}, \quad x < 1$$
2. 2

   Differentiating does not enlarge the set of allowed $x$ values. The derivative is defined over exactly the same open interval as the composite itself.

   $$\frac{d}{dx}(f \circ g)(x) = \frac{-1}{2(1 - x)}, \quad x \in (-\infty, 1)$$
3. 3

   So the domain of the derivative is the open interval $(-\infty, 1)$, closed at neither end.

   $$\text{dom} = (-\infty, 1)$$

VCAA 2023 Mathematical Methods Exam 2, Section A Q7

Consider $f : (1, \infty) \to R$, $f(x) = x^2 - 4x$ and $g : R \to R$, $g(x) = e^{-x}$. Find the range of the composite function $g(f(x))$.

1. 1

   First find the range of the inner function $f$ on $(1, \infty)$. The vertex is at $x = 2$, giving the minimum $f(2) = -4$, and $f \to \infty$ as $x$ grows.

   $$\text{ran}(f) = [-4, \infty)$$
2. 2

   Now apply the outer function $g(u) = e^{-u}$ to that range. Since $g$ is decreasing, the maximum occurs at the smallest input $u = -4$, giving $e^{4}$, while as $u \to \infty$, $e^{-u} \to 0^+$.

   $$\text{ran}(g \circ f) = (0, e^4]$$
3. 3

   The maximum $e^4$ is attained at $u = -4$ so the bracket is closed there, while $0$ is approached but never reached so it stays open.

   $$\text{ran}(g(f(x))) = (0, e^4]$$

VCAA 2024 Mathematical Methods Exam 2, Section A Q5

Given $f(x) = \sin(x)$ and $g(x) = \sin(2x)$. (i) The graph of $y = (g \circ f)(x)$ has a local maximum whose $x$-value lies in $\left[0, \dfrac{\pi}{2}\right]$. Find the coordinates of this local maximum, correct to one decimal place. (ii) State the range of $g \circ f$ where $x \in [0, 2\pi]$.

1. 1

   Build the composite by feeding $f$ into $g$. $(g \circ f)(x) = \sin(2 \sin x)$.

   $$(g \circ f)(x) = \sin(2 \sin x)$$
2. 2

   (i) The maximum value of $\sin$ is $1$, reached when its input is $\tfrac{\pi}{2}$. So set $2 \sin x = \tfrac{\pi}{2}$, giving $x = \sin^{-1}\!\left(\tfrac{\pi}{4}\right) \approx 0.9$, where $y = 1$.

   $$x = \sin^{-1}\!\left(\tfrac{\pi}{4}\right) \approx 0.9, \ y = 1$$
3. 3

   (ii) For $x \in [0, 2\pi]$, the inner expression $2 \sin x$ ranges over $[-2, 2]$, which includes $\pm \tfrac{\pi}{2}$, so $\sin(2 \sin x)$ attains every value in $[-1, 1]$.

   $$\text{range} = [-1, 1]$$

VCAA 2024 Mathematical Methods Exam 2, Section B Q5a

Let $f(x) = x^2 + 1$ and $g(x) = 2x - 3$. Find $f(g(2))$.

1. 1

   Work from the inside out. Do the inner machine $g$ first, at $x = 2$.

   $$g(2) = 2(2) - 3 = 1$$
2. 2

   Now feed that answer into the outer machine $f$.

   $$f(g(2)) = f(1) = 1^2 + 1 = 2$$

Let $f(x) = x^2 + 1$ and $g(x) = 2x - 3$. Find a simplified rule for $f(g(x))$.

1. 1

   Replace every $x$ in $f$ with the whole expression $g(x)$. Keep the bracket.

   $$f(g(x)) = (g(x))^2 + 1 = (2x - 3)^2 + 1$$
2. 2

   Expand $(2x-3)^2$ carefully. The middle term is the one students drop.

   $$(2x - 3)^2 = 4x^2 - 12x + 9$$
3. 3

   Add the $+1$ and collect like terms.

   $$f(g(x)) = 4x^2 - 12x + 10$$

Let $f(x) = \sqrt{x}$ with domain $[0, \infty)$ and $g(x) = x - 5$ with domain $R$. Is $f(g(x))$ defined? If so, give its rule and domain.

1. 1

   The composite $f(g(x))$ exists only if the range of the inner $g$ sits inside the domain of the outer $f$.

   $$\text{ran}(g) = R, \quad \text{dom}(f) = [0, \infty)$$
2. 2

   $R$ is not a subset of $[0, \infty)$, so $f(g(x))$ is not defined on all of $R$. Restrict $g$ so its outputs are at least $0$, that is $x - 5 \ge 0$.

   $$x \ge 5$$
3. 3

   On the restricted domain the rule is $f(g(x)) = \sqrt{g(x)}$.

   $$f(g(x)) = \sqrt{x - 5}, \quad x \in [5, \infty)$$