Composite and inverse functions: existence, formulas, domains and graphs

Composition is "outside applied to inside".

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

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

To invert $f$, swap $x$ and $y$ and solve: $x = 2 y - 3 \implies y = \frac{x + 3}{2}$.

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

Markers reward correct composition in both orders (they differ), the swap-and-solve method for the inverse, and a tidy final expression.

Restrict to $x \ge 0$. On this domain, $f$ is increasing and one-to-one, with range $[0, \infty)$.

Inverse: swap $x$ and $y$ in $y = x^2$ to get $x = y^2$ with $y \ge 0$, so $y = \sqrt{x}$.

$f^{-1}(x) = \sqrt{x}$ on domain $[0, \infty)$ with range $[0, \infty)$.

Markers expect a domain choice that yields a one-to-one function, a correct inverse derived by swap-and-solve with the right sign, and a domain and range that match.

Substitute the inner function into the outer one, both ways. For $f(g(x))$, feed $g(x) = x^2$ into $f$:

For $g(f(x))$, feed $f(x) = 3x + 1$ into $g$:

Marker's note: one mark for $f(g(x)) = 3x^2 + 1$, one mark for $g(f(x)) = 9x^2 + 6x + 1$ correctly expanded. Writing the same answer for both (assuming composition is commutative) loses a mark.

Swap $x$ and $y$, then solve. Write $y = 4x - 7$ and swap the variables:

So $f^{-1}(x) = \dfrac{x + 7}{4}$.

Verify by composition. Substitute $f^{-1}(x)$ into $f$:

Marker's note: one mark for the swap-and-solve method, one for $f^{-1}(x) = \frac{x + 7}{4}$, one for the verification simplifying to $x$. A verification that stops before the final $= x$ does not earn the last mark.

Substitute $g$ into $f$.

Apply both domain conditions. First $x$ must be in the domain of $g$, which is all of $\mathbb{R}$. Second, $g(x)$ must be in the domain of $f$, so $g(x) \ge 2$:

The natural domain of $f \circ g$ is $(-\infty, -1] \cup [1, \infty)$.

Marker's note: one mark for $\sqrt{x^2 - 1}$, one for setting up $g(x) \ge 2$, one for the domain $x \le -1$ or $x \ge 1$. Reading the domain straight off $\sqrt{x^2 - 1} \ge 0$ (giving the same answer here by luck) misses the required $g(x) \in \text{dom}(f)$ reasoning and is penalised when the two disagree.

Part (a): find the range. Since $e^{2x} > 0$ for every real $x$, we have $e^{2x} - 3 > -3$. As $x \to \infty$, $f(x) \to \infty$, and as $x \to -\infty$, $f(x) \to -3$ from above. So the range is $(-3, \infty)$.

Part (b): invert by swap-and-solve. Write $y = e^{2x} - 3$ and swap:

Take natural logs and solve for $y$:

So $f^{-1}(x) = \dfrac{1}{2}\ln(x + 3)$. Its domain is the range of $f$, namely $(-3, \infty)$.

Marker's note: one mark for the range $(-3, \infty)$ in (a); one for isolating $e^{2y} = x + 3$, one for $f^{-1}(x) = \frac{1}{2}\ln(x + 3)$, one for the domain $(-3, \infty)$ in (b). Forgetting the factor of $\frac{1}{2}$ (writing $\ln(x + 3)$) is the common slip.

Feature 1: the points are reflected in $y = x$. The first curve passes through $(0, 1)$ and $(1, 4)$. Reflecting each point in $y = x$ swaps its coordinates, giving $(1, 0)$ and $(4, 1)$, which are exactly the two points the second curve passes through. Every point $(a, b)$ on the first curve corresponds to $(b, a)$ on the second.

Feature 2: the asymptote swaps axes. The first curve has the $x$-axis (a horizontal asymptote); under reflection in $y = x$ a horizontal asymptote becomes a vertical one, and indeed the second curve has the $y$-axis as its asymptote. This is the graphical signature of domain and range swapping.

Because the two curves are mirror images in $y = x$ (matching swapped points and swapped asymptotes), the second is the inverse of the first.

Marker's note: one mark for identifying the swapped points $(0,1)\!\to\!(1,0)$ and $(1,4)\!\to\!(4,1)$, one for the horizontal-to-vertical asymptote swap, one for the conclusion tying both to reflection in $y = x$. A vague "they look symmetrical" without naming specific swapped features caps the answer at one mark.

Part (a): compute $f(f(x))$. The outer rule is $f(u) = \dfrac{3u + 1}{u - 3}$; here the input is $u = f(x) = \dfrac{3x + 1}{x - 3}$. Build the numerator and denominator separately, clearing the inner fraction by writing each over $(x - 3)$.

Numerator $3u + 1$:

Denominator $u - 3$:

Divide; the common factor $(x - 3)$ cancels:

Since $f(f(x)) = x$, the function undoes itself, so $f = f^{-1}$.

Part (b): meet the line $y = x$. Because $f$ is its own inverse, its graph is symmetric in $y = x$, and any intersection with that line solves $f(x) = x$:

Solve with the quadratic formula:

Both values satisfy $x \ne 3$, so the graph meets $y = x$ at $\left(3 + \sqrt{10}, \, 3 + \sqrt{10}\right)$ and $\left(3 - \sqrt{10}, \, 3 - \sqrt{10}\right)$.

Marker's note: two marks for correctly building and simplifying $f(f(x))$ to $x$ (writing each of $3u + 1$ and $u - 3$ over $x - 3$ is the key step), one for stating $f = f^{-1}$; one for reducing $f(x) = x$ to $x^2 - 6x - 1 = 0$, one for both exact points $3 \pm \sqrt{10}$. Leaving the intersection as a decimal without the exact surd form is penalised.

Part (a): show one-to-one on $x \ge 2$. Complete the square:

The vertex is at $x = 2$. For $x \ge 2$ the factor $(x - 2)^2$ is strictly increasing, so $f$ is strictly increasing on this domain. A strictly increasing function passes the horizontal line test, so it is one-to-one and an inverse exists.

Part (b): invert on the restricted branch. From $y = (x - 2)^2 + 3$, swap $x$ and $y$:

Take the square root. Because the restriction $x \ge 2$ forces $y \ge 2$ on the inverse (the range endpoint carries over), choose the positive root:

So $f^{-1}(x) = 2 + \sqrt{x - 3}$. The domain of $f^{-1}$ is the range of $f$; since the minimum value of $f$ on $x \ge 2$ is $f(2) = 3$, the range is $[3, \infty)$, so the domain of $f^{-1}$ is $[3, \infty)$.

Marker's note: one mark for completing the square and arguing strictly increasing (hence one-to-one) in (a); one for reaching $(y - 2)^2 = x - 3$, one for choosing $+\sqrt{\,}$ with a reason and writing $f^{-1}(x) = 2 + \sqrt{x - 3}$, one for the domain $[3, \infty)$ in (b). Taking $-\sqrt{\,}$ (which suits the $x \le 2$ branch instead) is the classic sign error.