Functions: past paper questions & worked solutions

(a) The denominator of \(g\) cannot be zero: \[ 2x+1=0 \]

Now \[ g(k)=3k+1 \]

So \[ f(g(k))=\frac{k}{2(3k+1)-4}=\frac{k}{6k-2} \]

Since the domain is \(x<4\), take the negative square root:

So \[ x<-\frac92 \quad \text{or} \quad x>\frac92 \]

If the input is restricted to \(x\ge0\), this becomes \(x>\frac92\).

$g(x) = \\dfrac{3x}{x-2}$. (a) Value excluded from domain. (b) Find *g*⁻¹(*x*).

The denominator must not equal zero: *x* − 2 ≠ 0.

*x* = 2 cannot be in any domain of *g*

Set *y* = *g*(*x*) and swap *x* ↔ *y*: $$x = \\frac{3y}{y-2}$$

Multiply both sides by (*y*−2): *x*(*y*−2) = 3*y* ⟹ *x**y* − 2*x* = 3*y*

Collect *y* terms and factorise: *x**y* − 3*y* = 2*x* ⟹ *y*(*x*−3) = 2*x*

Solve for *y*: $$y = \\frac{2x}{x-3}$$

*g*⁻¹(*x*) is undefined at *x* = 3, which is the range restriction of *g*(*x*) (horizontal asymptote).

Complete the square:

Let \(y=f(x)\).

Since \(x\ge 6\), take the positive square root:

Replace \(y\) by \(x\).

(a) The denominator cannot be zero, so \[ x-2=0 \]

So \[ f^{-1}(x)=\frac{2x+1}{x-3} \]

(a) The denominator of \(g(x)\) cannot be zero.

So \(x=\frac52\) must be excluded.

Since \(x\ge0\), \[ x=2 \]

So \[ f^{-1}(x)=\frac{6x}{x-2} \]

Factorise: \[ (2x+3)(2x-1)>0 \]

The product is positive outside the roots.

Since the domain of \(h\) is \(x<6\), take the negative square root:

(b) A square is always greater than or equal to zero, so the range is \[ f(x)\ge0 \]

For the inverse, let

So

Substitute 4 for the function input.

If \(y=4+x\), then \(x=y-4\), so \(f^{-1}(x)=x-4\).

Reverse the input-output relationship.

Apply the inverse function to 7.

Apply f first and use its whole output as the input of g.

Expand and collect the constant terms.

Complete the square:

The domain of \(g\) is \(x\ge 3\), so take the positive square root:

Therefore

The range of \(g\) is \(x\le 14\), so this is the domain of \(g^{-1}\).

Since the domain is \(x\ge -2\), take the positive square root:

(a) The denominator of \(g\) cannot be zero: \[ 3x+1=0 \]

So \[ g^{-1}(x)=\frac{x}{1-3x} \]

Since the domain is \(x\ge 1\), take the positive square root:

For \(f(x)=\frac1x\), the denominator is zero when \(x=0\).

Division by zero is undefined, so this input must be excluded.

In gf, apply f first.

The input 2 lies in the stated domain x>=1 of g.

Complete the square to isolate the expression containing x.

Choose the nonnegative root because the domain of g is x>=1.

Rearrange to make the original input the subject.

Exchange the input and output labels; the range of g becomes the domain of the inverse.

**(a) Find *g*(−1):** *g*(−1) = 4(−1)² + 16(−1) − 9 = 4 − 16 − 9 =  − 21

\(a\)  − 21

**(b) Find *f**g*(*x*):** *f**g*(*x*) = *f*(*g*(*x*)) = 5(4*x*²+16*x*−9) − 3  = 20*x*² + 80*x* − 45 − 3 = 20*x*² + 80*x* − 48

\(b\) 20*x*² + 80*x* − 48

Let *y* = 4*x*² + 16*x* − 9. We must rearrange to make *x* the subject.

**Step 1: Complete the square for the *x* terms.** *y* = 4(*x*²+4*x*) − 9 *y* = 4((*x*+2)²−4) − 9 *y* = 4(*x*+2)² − 16 − 9 *y* = 4(*x*+2)² − 25

**Step 2: Rearrange for *x*.** *y* + 25 = 4(*x*+2)² $$\\frac{y + 25}{4} = (x + 2)^2$$ $$x + 2 = \\pm\\sqrt{\\frac{y + 25}{4}}$$

Since *x* ≥  − 2, *x* + 2 is positive, so we take the positive root. $$x + 2 = \\frac{\\sqrt{y + 25}}{2}$$ $$x = \\frac{\\sqrt{y + 25}}{2} - 2$$

**Step 3: Swap *y* for *x* to write the inverse function.**

(a) If \(g(g(x))=6\), then

So

Let \(y=5(x+3)^2-52\). Since \(x\ge -3\), use the positive square root:

(a) The denominator of \(g\) cannot be zero: \[ 2x-9=0 \]

Complete the square: \[ h(x)=3(x-2)^2-4 \]

Since \(x>2\), take the positive root: \[ x=2+\sqrt{\frac{y+4}{3}} \]

So \[ h^{-1}(x)=2+\sqrt{\frac{x+4}{3}} \]

Solve \(fg(x)=f(x)\): \[ (x-10)^2+6=x^2+6 \]

The denominator cannot be zero, so \(x=0\) cannot be included.

Therefore \[ h^{-1}(x)=\frac{4}{2-x} \]

(a) The denominator of \(g\) cannot be zero, so \[ x=-4 \]

So \[ g^{-1}(x)=\frac{4x}{5-x} \]