Review question

# Can we find an inverse and sketch a modulus function? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5646

## Solution

1. A function is defined by $f:x\mapsto \dfrac{1}{3-2x}$ for all values of $x$ except $x=\dfrac{3}{2}$.

1. Find the values of $x$ which map on to themselves under the function $f$.

We seek $x$ such that $f(x)=x$, so we need to solve $x=\frac{1}{3-2x}.$ Multiplying both sides by $3-2x$, we find that $3x-2x^2=1,$ so we must solve the quadratic equation $2x^2-3x+1=0.$ By inspection, we see that $2x^2-3x+1=(2x-1)(x-1),$ therefore the points which are mapped to themselves by $f$ are $x=\dfrac{1}{2}$ and $x=1$.

1. A function is defined by $f:x\mapsto \dfrac{1}{3-2x}$ for all values of $x$ except $x=\dfrac{3}{2}$.

1. Find an expression for $f^{-1}$.
Write $y=\dfrac{1}{3-2x}$. If we can write $x$ in terms of $y$ then we have found $f^{-1}$. We have that $3y-2xy=1,$ and so $2xy=3y-1,$ therefore $x=\frac{3y-1}{2y}.$ So we have that $f^{-1}(x)=\frac{3x-1}{2x}.$
1. A function is defined by $f:x\mapsto \dfrac{1}{3-2x}$ for all values of $x$ except $x=\dfrac{3}{2}$.

1. Evaluate $ff(2)$.

We have that $f(2)=\dfrac{1}{3-4}=-1$, so $ff(2)=f(-1)=\dfrac{1}{3+2}=\dfrac{1}{5}$.

1. Sketch the graph of $y=|x^2-4x+3|$ for the domain $0\le x \le 5$, and state the corresponding range of $y$.

To start with, let’s consider $x^2-4x+3=(x-1)(x-3)$. Using our knowledge of quadratics, we can sketch this straightforwardly.

When $x=0$, $x^2-4x+3=3$, and when $x=5$, $x^2-4x+3=8$. The minimum occurs at the midpoint of the two roots, when $x=2$, and here $x^2-4x+3=-1$.

To make this into a sketch of $y=|x^2-4x+3|$, we must reflect the part where $x^2-4x+3<0$ in the $x$-axis.

The largest value of $y$ is $8$ when $x=5$. Therefore for the domain $0\le x\le 5$, we have that the range of $y=|x^2-4x+3|$ is $0\le y\le 8$.