Functions: Question 8

Worked solution

Part (a): Completing the square and the range of $f$

$f(x) = x^2 - 6x + 11$

Halve the coefficient of $x$ ($-6 \div 2 = -3$) and complete the square:

$x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2$

So $f(x) = (x-3)^2+2$. This is an upward-opening parabola with vertex $(3,2)$.

Since the domain is restricted to $x \ge 3$, we only see the increasing branch starting at the vertex. The smallest value of $f$ occurs at $x=3$, where $f(3) = (3-3)^2+2 = 2$, and $f$ increases without bound as $x$ increases. So the range is: $f(x) \ge 2$

Part (b): Finding $f^{-1}(x)$

Write $y = f(x)$ using the completed-square form:

$y = (x-3)^2 + 2, \qquad x \ge 3$

Rearrange for $x$ in terms of $y$. Subtract $2$:

$y - 2 = (x-3)^2$

Take square roots. Since the domain is $x\ge3$, we have $x-3\ge0$, so we must take the positive square root:

$\sqrt{y-2} = x - 3$

$x = 3 + \sqrt{y-2}$

Now swap $x$ and $y$ to write the inverse as a function of $x$:

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

Part (c): Domain of $f^{-1}$

The domain of $f^{-1}$ always equals the range of $f$, found in part (a):

$x \in \mathbb{R}, \quad x \ge 2$

Part (d): Solving $f^{-1}(x) = 7$

Rather than substituting into the formula directly, use the key inverse-function fact: if $f^{-1}(x) = 7$, then applying $f$ to both sides gives $x = f(7)$.

$f(7) = (7-3)^2 + 2 = 16 + 2 = 18$

So $x = 18$.

Check using the formula from (b): $f^{-1}(18) = 3 + \sqrt{18-2} = 3+\sqrt{16} = 3+4 = 7$ ✓, confirming $f^{-1}(18)=7$, as required.

Final answers

• (a) $f(x) = (x-3)^2+2$; range: $f(x) \ge 2$
• (b) $f^{-1}(x) = 3+\sqrt{x-2}$
• (c) Domain of $f^{-1}$: $x \ge 2$
• (d) $x = \boxed{18}$