P3.1 - Algebra and functions

A rational expression is simplified by factorising numerator and denominator, cancelling common factors, and using algebraic division when the numerator's degree is at least the denominator's. Record excluded denominator values before cancelling.

For the specification example, $$\frac{x^3+1}{x^2-1}=\frac{(x+1)(x^2-x+1)}{(x+1)(x-1)}=\frac{x^2-x+1}{x-1},\qquad x\ne-1,1.$$ Cancelling removes a common factor, not the original restriction $x\ne-1$.

Now divide: $x^2-x+1=x(x-1)+1$, so $$\frac{x^3+1}{x^2-1}=x+\frac1{x-1},\qquad x\ne-1,1.$$ The quotient has lower-degree remainder, which is the useful stopping form.

For an unknown quotient and remainder, write an identity such as $$\frac{4x^3+2x^2+3x+8}{x^2+4}=Ax+B+\frac{Cx+D}{x^2+4},$$ multiply through by the denominator, then equate coefficients or substitute convenient values.

A function assigns exactly one output to each input in its domain. Different inputs may share an output, so a many-one mapping is still a function; an inverse function exists only after the mapping is one-one on its chosen domain.

Let $f(x)=2x+3$ on $\mathbb R$ and $g(x)=x^2$ for $x\ge0$. Then $$fg(x)=f(g(x))=2x^2+3,\qquad x\ge0.$$ The restriction makes $g$ one-one, so $g^{-1}(x)=\sqrt{x}$ for $x\ge0$. Domain and range exchange when a function is inverted.

The graphs of $y=f(x)$ and $y=f^{-1}(x)$ are reflections in $y=x$. If a horizontal line meets the original graph more than once, restrict its domain before claiming an inverse function.

Modulus makes a quantity non-negative: $|u|=u$ for $u\ge0$ and $|u|=-u$ for $u<0$. This piecewise rule controls both sketches and algebraic solutions.

For $|2x-1|=x+5$, use the two linear branches: $$2x-1=x+5\Rightarrow x=6,$$ $$-(2x-1)=x+5\Rightarrow x=-\frac43.$$ Both satisfy their corresponding sign condition, so both are solutions.

The same expressions meet at $x=-\frac43$ and $x=6$. Testing the intervals, or comparing the two graphs, gives $$|2x-1|>x+5\quad\Longrightarrow\quad x<-\frac43\text{ or }x>6.$$ Strict inequality excludes the intersection points.

Transformations outside f change output coordinates; transformations inside f change input coordinates in the opposite way. Map a known point rather than relying on a verbal left/right guess.

A negative outside factor also reflects in the x-axis; a negative inside factor reflects in the y-axis. Magnitudes greater than 1 stretch vertically outside but compress horizontally inside.

If $(6,-2)$ lies on $y=f(x)$, then on $$y=2f(3x)+5$$ the input coordinate becomes $6/3=2$ and the output becomes $2(-2)+5=1$. Thus the mapped point is $(2,1)$.

For a permitted combination, apply every horizontal coordinate change to u and every vertical change to v, then use intercepts and mapped features to complete the sketch. For example, $y=f(-x)+1$ maps $(u,v)$ to $(-u,v+1)$.