The Properties of the Tangent Function

Determine the properties of the function $f(x)=-\tan\left(\frac{1}{2}(x-1)\right)+\sqrt{2}$ .

It may be useful to plot a graph of the function.

• The coordinates of the inflection point are $(h,k)=(1,\sqrt{2})$.

• The period of the function is: $$\displaystyle P = \frac{\pi}{\mid b \mid} = \frac{\pi}{\frac{1}{2}} = 2\pi$$

• The equation of the asymptotes are: $$\begin{align} x &= \left(h + \frac{P}{2}\right) + n P\\ &= \left(1+\frac{2\pi}{2}\right) +n (2\pi)\\ &= (1+\pi) + 2\pi n \end{align}$$ where $n \in \mathbb{Z}$ and $P$ is the period.

• The domain of the function is: $\mathbb{R} \backslash \lbrace (1+\pi) + 2 \pi n \rbrace$ where $n \in \mathbb{Z}$ and $P$ is the period.

• The range of the function is the set of real numbers, i.e. $\mathbb{R}.$

• The interval: from the values of $a$ and $b,$ the function must be decreasing, since the product $a b$ is negative $\left(-1 \times \frac{1}{2} <0\right).$ The graph confirms it. 

• The zeroes of the function are calculated by replacing $f(x)$ by $0.$ $$\begin{align}0 &= -\tan\left(\frac{1}{2}(x-1)\right)+\sqrt{2}\\-\sqrt{2} &= - \tan\left(\frac{1}{2}(x-1)\right)\\ \sqrt{2} &= \tan\left(\frac{1}{2}(x-1)\right)\end{align}$$ At this step, check what angle the tangent is $\sqrt{2}.$ Look at the interval angle $[0,\pi].$ The value is $0{.}955.$
  
  Thus, the interior of the tangent function is equal to $0{.}955.$ $$\begin{align}0{.}955 &= \frac{1}{2}(x-1)\\1{.}91 &= x-1\\2{.}91 &= x\end{align}$$ The zero of the function in the cycle is $2{.}91.$
  
  The general expression for the function’s zeroes is $x=2{.}91 + 2\pi n$ where $n \in \mathbb{Z}.$

• The positive and negative intervals: the function is positive on the interval $(1-\pi + 2\pi n,\ 2{.}91 + 2 \pi n]$ and negative on the interval $[2{.}91 + 2 \pi n,\ 1+ \pi + 2 \pi n)$ where $n \in \mathbb{Z}$.
  Be careful not to include asymptotes.

• The function has no extrema.