Composite Functions

Apply the function nearest the input first.

1. First find g(6)=6-4=2.
2. The flow makes the order clear: apply g to 6 first, then apply f to the result.

3. Then f(2)=2\times2+1=5, so fg(6)=5.

For this order, apply the square function first.

1. First f(2)=4.
2. Then g(4)=4+3=7, so gf(2)=7.

Work from the input towards the outside function.

1. f(-2)=5-(-2)=7.
2. Then g(7)=21, so gf(-2)=21.

Use the output of the second function as the denominator in the first.

1. g(5)=3.
2. Then f(3)=\frac{6}{3}=2.

Replace the input in the outside rule by the whole inside rule.

1. fg(x)=f(x^2).
2. Replacing the input in 2x-3 gives 2x^2-3.

The outside function adds to the output of the inside function.

1. gf(x)=g(x^2-1).
2. This is x^2-1+4, which simplifies to x^2+3.

Substitute the complete linear expression into the outside function.

1. fg(x)=3(2x-5)+2.
2. Expand and collect to get 6x-13.

Square the whole inside expression.

1. fg(x)=(2x+1)^2.
2. Expanding the squared bracket gives 4x^2+4x+1.

Applying the outside function divides the inside output by two.

1. fg(x) means put g(x) into f: fg(x)=f\left(\frac{6}{x}\right).
2. f halves its input: \frac{6}{x}\div2.
3. \frac{6}{x}\div2=\frac{6}{2x}=\frac{3}{x}.

ff(x) means f(f(x)): put the whole of 2x-5 into f.

1. ff(x) means apply f twice: ff(x)=f(2x-5).
2. Replace x in 2x-5 by (2x-5): 2(2x-5)-5.
3. Expand and collect: 4x-10-5=4x-15.

Find both composite rules before comparing them.

1. fg(x) means apply g first: fg(x)=f(x+4)=2(x+4)+1=2x+9.
2. gf(x) means apply f first: gf(x)=g(2x+1)=(2x+1)+4=2x+5.
3. The two flows show why the order matters: fg(x)=2x+9, while gf(x)=2x+5.

4. 2x+9 is always 4 more than 2x+5, so fg(x)=gf(x)+4.
5. The first option is wrong: the order matters, and here fg(x) and gf(x) are different.

Write the composite as an ordinary equation in the input.

1. fg(x)=3(x-2)+1.
2. Solve 3x-5=16, giving 3x=21.
3. Therefore x=7.

Form the composite rule, then remember both square roots.

1. gf(x)=x^2+2.
2. The equation gives x^2=9.
3. Therefore x=-3 or x=3.

Find the input that would make the composite denominator zero.

1. fg(x)=\frac{1}{2x+1-3}.
2. The input to f is g(x)=2x+1. The denominator of f(u) is zero when that input is 3.

3. Its denominator is zero when 2x-2=0.
4. The excluded input is x=1.

The whole inside expression is squared before the final constant is added.

1. fg(x)=(3-x)^2+1.
2. Expanding gives 9-6x+x^2+1.
3. Collect terms: x^2-6x+10.

Form both composite expressions and put all terms on one side.

1. fg(x) means apply g first: fg(x)=f(x^2)=2x^2+1.
2. gf(x) means apply f first: gf(x)=g(2x+1)=(2x+1)^2=4x^2+4x+1.
3. Set them equal: 2x^2+1=4x^2+4x+1.
4. Rearrange: 0=2x^2+4x, and factorise: 2x(x+2)=0.
5. So x=0 or x=-2.

Combine the terms in the inner denominator before simplifying the outer fraction.

1. fg(x) means put g(x) into f: fg(x)=\frac{4}{\frac{x-3}{2}+1}.
2. Simplify the denominator: \frac{x-3}{2}+\frac{2}{2}=\frac{x-1}{2}.
3. Dividing by \frac{x-1}{2} is the same as multiplying by \frac{2}{x-1}: 4\times\frac{2}{x-1}=\frac{8}{x-1}.

Write h(x) as a single expression first, then make x the subject of y=h(x).

1. fg(x) means apply g first, then f: h(x)=f(x+3)=\frac{12}{x+3}.
2. To find the inverse, write y=\frac{12}{x+3} and make x the subject.
3. Multiply both sides by x+3: y(x+3)=12, so x+3=\frac{12}{y}.
4. Subtract 3: x=\frac{12}{y}-3.
5. Write the inverse in terms of x: h^{-1}(x)=\frac{12}{x}-3, which can also be written as \frac{12-3x}{x}.

Isolate the squared bracket, then use both signs of the square root.

1. fg(x)=(2x+3)^2-2.
2. The equation gives (2x+3)^2=25.
3. So 2x+3=5 or 2x+3=-5.
4. Therefore x=1 or x=-4.

Expand the composite rule and compare the coefficient and constant with the stated result.

Part (a)

1. gf(x)=a(3x-2)+b=3ax-2a+b.
2. Compare coefficients: 3a=6, so a=2.

Part (b)

1. The constants satisfy -2a+b=5.
2. Using the value of a gives -4+b=5, so b=9.