Inverse Functions

Reverse the operations in the opposite order.

1. Write y=3x+4 and make x the subject.
2. The inverse reverses the forward operations in reverse order: subtract 4, then divide by 3.

3. Subtract 4 from both sides: y-4=3x.
4. Divide both sides by 3: x=\frac{y-4}{3}.
5. This rule takes an output of f back to its input. Write it with x as the input: f^{-1}(x)=\frac{x-4}{3}.

The inverse must undo the subtraction.

1. The original rule subtracts 7.
2. The inverse adds 7, so f^{-1}(x)=x+7.

Use an equation to keep track of the negative coefficient.

1. Write y=5-2x and make x the subject.
2. Add 2x to both sides and subtract y: 2x=5-y.
3. Divide by 2: x=\frac{5-y}{2}.
4. Replace y with x: f^{-1}(x)=\frac{5-x}{2}.

Subtract before undoing the division.

1. Write y=\frac{x}{4}+6 and make x the subject.
2. Subtract 6: y-6=\frac{x}{4}.
3. Multiply by 4: x=4(y-6).
4. Replace y with x and expand the bracket: f^{-1}(x)=4x-24.

Undo the denominator before undoing the numerator operations.

1. Write y=\frac{2x+1}{3} and make x the subject.
2. Multiply both sides by 3: 3y=2x+1.
3. Subtract 1, then divide by 2: x=\frac{3y-1}{2}.
4. Replace y with x: f^{-1}(x)=\frac{3x-1}{2}.

Find the input that would produce the stated output.

1. g^{-1}(12) is the input that g turns into 12, so solve g(x)=12.
2. 7x-9=12, so 7x=21.
3. x=3, so g^{-1}(12)=3.

Use the given restriction to choose the appropriate square root.

1. Write y=x^2. Square rooting gives x=\sqrt{y} or x=-\sqrt{y}.
2. The inputs of f are x\ge0, so the inverse must give back a value of 0 or more: take the positive root, x=\sqrt{y}.
3. Because the original input is non-negative, its inverse uses the non-negative square root.

4. Replace y with x: f^{-1}(x)=\sqrt{x}.

Find the input x for which f(x)=31.

1. f^{-1}(31) is the input that f turns into 31, so solve \frac{x^3}{2}-1=31.
2. Add 1: \frac{x^3}{2}=32.
3. Multiply by 2: x^3=64.
4. Take the cube root: x=4, so f^{-1}(31)=4.

Undo the addition before undoing the square.

1. Write y=x^2+4 and make x the subject.
2. Subtract 4: x^2=y-4.
3. The inputs of f are x\ge0, so take the positive square root: x=\sqrt{y-4}.
4. Replace y with x: f^{-1}(x)=\sqrt{x-4}.

Clear the denominator and solve for the original input.

1. Write y=\frac{3}{x-2} and multiply both sides by x-2: y(x-2)=3.
2. Expand the bracket: xy-2y=3.
3. Add 2y to both sides: xy=2y+3.
4. Divide by y: x=\frac{2y+3}{y}.
5. Replace y with x: f^{-1}(x)=\frac{2x+3}{x}.

Work from the inside out: the function next to the bracket is used first.

1. gf^{-1}(11) means g\left(f^{-1}(11)\right): use f^{-1} first, then g.
2. f^{-1}(11) is the input that f turns into 11: 2x+3=11, so x=4.
3. Now use g: g(4)=4^2=16.
4. So gf^{-1}(11)=16.

Apply the original function to both sides of the equation.

1. f^{-1}(x)=5 means that f turns 5 into x, so x=f(5).
2. f(5)=4\times5-7=13, so x=13.
3. Check with the inverse: f^{-1}(x)=\frac{x+7}{4}, and \frac{13+7}{4}=5.

Take the root that makes the original input satisfy its restriction.

1. Write y=(x-1)^2. Square rooting gives x-1=\sqrt{y} or x-1=-\sqrt{y}.
2. The inputs of f satisfy x\ge1, so x-1 is 0 or more. Take the positive root: x-1=\sqrt{y}.
3. Since x\ge1, the quantity x-1 is non-negative, so the inverse takes the positive square root before adding 1.

4. Add 1: x=1+\sqrt{y}.
5. Replace y with x: f^{-1}(x)=1+\sqrt{x}.

Square the output before undoing the addition inside the root.

1. Write y=\sqrt{x+2} and make x the subject.
2. Square both sides to undo the square root: y^2=x+2.
3. Subtract 2: x=y^2-2.
4. Replace y with x: f^{-1}(x)=x^2-2.

Find f^{-1}(x) first, then set the two expressions equal to each other.

1. Find the inverse: write y=3x-8, so x=\frac{y+8}{3} and f^{-1}(x)=\frac{x+8}{3}.
2. Set the two expressions equal: 3x-8=\frac{x+8}{3}.
3. Multiply both sides by 3: 9x-24=x+8.
4. Collect terms: 8x=32, so x=4.
5. Check: f(4)=4 and f^{-1}(4)=\frac{12}{3}=4.

After clearing the fraction, collect all terms containing the original input.

1. Write y=\frac{2x-3}{x+4} and multiply both sides by x+4: y(x+4)=2x-3.
2. Expand: xy+4y=2x-3.
3. Get the terms with x on one side: 4y+3=2x-xy.
4. Factorise the right-hand side: 4y+3=x(2-y).
5. Divide by 2-y: x=\frac{4y+3}{2-y}, so f^{-1}(x)=\frac{4x+3}{2-x}.

Make x^2 the subject, then take the positive square root, since x\ge0.

1. Write y=2x^2-5 and make x the subject.
2. Add 5, then divide by 2: x^2=\frac{y+5}{2}.
3. The inputs of f are x\ge0, so take the positive square root: x=\sqrt{\frac{y+5}{2}}.
4. Replace y with x: f^{-1}(x)=\sqrt{\frac{x+5}{2}}.

Find f^{-1}(x), set it equal to g(x), then clear the fraction.

1. Find the inverse: write y=3x+1, so x=\frac{y-1}{3} and f^{-1}(x)=\frac{x-1}{3}.
2. Set it equal to g(x): \frac{x-1}{3}=x^2-1.
3. Multiply both sides by 3: x-1=3x^2-3.
4. Rearrange so one side is 0: 3x^2-x-2=0.
5. Factorise: (3x+2)(x-1)=0, so x=1 or x=-\frac{2}{3}.

Reverse the given inverse rule to recover the original function.

Part (a)

1. The inverse of f^{-1} is f itself, so find the inverse of y=\frac{x+7}{4}.
2. Multiply by 4: 4y=x+7, so x=4y-7.
3. So f(x)=4x-7. Comparing with f(x)=ax+b gives a=4.

Part (b)

1. From part (a), f(x)=4x-7, so b=-7.
2. Check: f(1)=-3 and f^{-1}(-3)=\frac{-3+7}{4}=1.

Undo the outside addition before squaring.

1. Write y=\sqrt{3x-2}+4 and make x the subject.
2. Subtract 4: y-4=\sqrt{3x-2}.
3. Square both sides: (y-4)^2=3x-2.
4. The inverse undoes the operations in reverse order. Its input restriction x\ge4 matches the range of the original function.

5. Add 2, then divide by 3: x=\frac{(y-4)^2+2}{3}.
6. Replace y with x: f^{-1}(x)=\frac{(x-4)^2+2}{3}.