Functions, Worked Examples (KBAT)

(a) $fg(x)=f(g(x))$, so substitute $g(x)=\dfrac{x+1}{2}$ into $f$:

$$fg(x)=3\!\left(\frac{x+1}{2}\right)-2=\frac{3x+3}{2}-2=\frac{3x+3-4}{2}=\frac{3x-1}{2}$$

To invert, let $y=\dfrac{3x-1}{2}$ and make $x$ the subject:

$$2y=3x-1 \;\Rightarrow\; 3x=2y+1 \;\Rightarrow\; x=\frac{2y+1}{3}$$

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

(b) Invert each function separately. For $f$, let $y=3x-2$:

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

For $g$, let $y=\dfrac{x+1}{2}$:

$$2y=x+1 \;\Rightarrow\; g^{-1}(x)=2x-1$$

Now build $g^{-1}f^{-1}(x)=g^{-1}\!\left(f^{-1}(x)\right)$ by feeding $f^{-1}(x)=\dfrac{x+2}{3}$ into $g^{-1}$:

$$g^{-1}f^{-1}(x)=2\!\left(\frac{x+2}{3}\right)-1=\frac{2x+4}{3}-1=\frac{2x+4-3}{3}=\frac{2x+1}{3}$$

(a) Complete the square. Half of $-6$ is $-3$, and $(-3)^{2}=9$:

$$f(x)=x^{2}-6x+5=(x-3)^{2}-9+5=(x-3)^{2}-4$$

(b) On the domain $x\geq 3$, the bracket $x-3\geq 0$, so $(x-3)^{2}\geq 0$ and takes every value from $0$ upward. Therefore $f(x)\geq -4$, with the minimum $-4$ reached at $x=3$.

$$\text{Range: } f(x)\geq -4$$

(c) To invert, set $y=(x-3)^{2}-4$ and make $x$ the subject:

$$(x-3)^{2}=y+4 \;\Rightarrow\; x-3=\pm\sqrt{y+4}$$

Because the domain is $x\geq 3$, we have $x-3\geq 0$, so take the positive square root only:

$$x=3+\sqrt{y+4} \;\Rightarrow\; f^{-1}(x)=3+\sqrt{x+4}$$

$f^{2}(x)=ff(x)=f(ax+b)$. Substitute the whole of $ax+b$ into $f$:

$$f^{2}(x)=a(ax+b)+b=a^{2}x+ab+b$$

Compare with $4x+9$. The coefficients of $x$ give $a^{2}=4$, and since $a>0$ we take the positive root:

$$a^{2}=4 \;\Rightarrow\; a=2 \quad (a>0)$$

Compare the constant terms, $ab+b=9$, factorise, and substitute $a=2$:

$$b(a+1)=9 \;\Rightarrow\; 3b=9 \;\Rightarrow\; b=3$$

So $f(x)=2x+3$. To find the inverse, let $y=2x+3$ and make $x$ the subject:

$$2x=y-3 \;\Rightarrow\; f^{-1}(x)=\frac{x-3}{2}$$

Since $fg(x)=f(g(x))$, the whole expression must be rewritten purely in terms of $g(x)$. Start by completing the square on the right-hand side:

$$4x^{2}+4x+3=(2x+1)^{2}+2$$

The bracket $(2x+1)^{2}$ is exactly $[g(x)]^{2}$, since $g(x)=2x+1$. Substitute:

$$fg(x)=[g(x)]^{2}+2$$

Comparing both sides, whatever is applied to $g(x)$ is the rule for $f$: square the input, then add $2$.

$$f(x)=x^{2}+2$$

Build $fg(x)=f(g(x))$ by substituting $g(x)=x^{2}-3$ into $f$:

$$fg(x)=\sqrt{(x^{2}-3)-2}=\sqrt{x^{2}-5}$$

This is only defined where the input to $f$, which is $g(x)$, satisfies $f$'s own condition $g(x)\geq 2$:

$$x^{2}-3\geq 2 \;\Rightarrow\; x^{2}\geq 5$$

Solve this inequality by taking square roots on both sides, keeping both branches since $x^{2}\geq 5$ allows $x$ to be large positive or large negative:

$$x\leq -\sqrt{5} \;\text{ or }\; x\geq \sqrt{5}$$

Let $y=f(x)$ and clear the fraction by multiplying both sides by $(x-3)$:

$$y=\frac{2x+1}{x-3} \;\Rightarrow\; y(x-3)=2x+1$$

Expand, then collect every term containing $x$ on one side:

$$xy-3y=2x+1 \;\Rightarrow\; xy-2x=3y+1$$

Factorise out $x$, then divide to make $x$ the subject:

$$x(y-2)=3y+1 \;\Rightarrow\; x=\frac{3y+1}{y-2}$$

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

Complete the square to locate the vertex. Half of $-4$ is $-2$, and $(-2)^{2}=4$:

$$f(x)=x^{2}-4x+1=(x-2)^{2}-4+1=(x-2)^{2}-3$$

The vertex is at $x=2$, which lies inside the given domain $0\leq x\leq 5$, so the minimum value $-3$ is actually reached, at $x=2$.

Because the domain is a closed interval rather than an unbounded one, the maximum is not automatic, it must be checked at both endpoints:

$$f(0)=(0-2)^{2}-3=1, \qquad f(5)=(5-2)^{2}-3=6$$

$x=5$ is farther from the vertex $x=2$ than $x=0$ is, so it gives the larger value. The maximum on this domain is $6$, at $x=5$.

$$\text{Range: } -3\leq f(x)\leq 6$$

First evaluate $g(3)$, since $fg(3)=f(g(3))$:

$$g(3)=3^{2}+1=10$$

Substitute into $f$ and set the result equal to $17$:

$$fg(3)=f(10)=2(10)-k=20-k=17 \;\Rightarrow\; k=3$$

So $f(x)=2x-3$. Now build $gf(x)=g(f(x))$ by substituting $f(x)=2x-3$ into $g$:

$$gf(x)=(2x-3)^{2}+1=4x^{2}-12x+9+1=4x^{2}-12x+10$$