Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\frac{1}{2}x-1$, then on the interval $[0,\pi ]$

We are examining the function $f(x)=\frac{1}{2}x-1$ defined on the closed interval $[0,\pi ]$. Our objective is to determine the continuity of the composite functions $\tan ([f(x)])$ and $\frac{1}{f(x)}$ within this domain.

First, we analyze the behavior of the base function $f(x)=\frac{1}{2}x-1$. At the lower bound $x=0$, the function value is $f(0)=-1$.

As $x$ increases to $\pi$, the function value reaches:

Thus, the range of $f(x)$ over the interval $[0,\pi ]$ is $[-1,\frac{\pi }{2}-1]$. This range serves as the input domain for our subsequent composite functions.

We now consider the function $g(x)=\frac{1}{f(x)}$. A rational function is discontinuous wherever its denominator equals zero.

Setting the denominator to zero:

Since $2$ lies within the interval $[0,\pi ]$, the function $\frac{1}{f(x)}$ encounters a vertical asymptote at $x=2$. Therefore, the function $\frac{1}{f(x)}$ is discontinuous at $x=2$.

Next, we examine $h(x)=\tan ([f(x)])$, where $[\cdot ]$ denotes the Greatest Integer Function. This function is discontinuous whenever the argument of the GIF is an integer.

In our range $[-1,0.57]$, the integers are $-1$ and $0$. We check the behavior at $x=2$, where $f(2)=0$:

1. As $x\to 2^{-}$, $f(x)<0$, so $[f(x)]=-1$. This yields $\tan (-1)$. 2. As $x\to 2^{+}$, $f(x)>0$, so $[f(x)]=0$. This yields $\tan (0)=0$.

Because $\tan(-1) eq 0$, the function exhibits a jump discontinuity at $x=2$. Consequently, the function $\tan ([f(x)])$ is discontinuous at $x=2$.

Through our analysis, we have determined that both functions fail to be continuous at the point $x=2$.

The reciprocal function $\frac{1}{f(x)}$ experiences an infinite discontinuity, while the composite function $\tan ([f(x)])$ experiences a jump discontinuity. Both points of failure occur within the specified interval $[0,\pi ]$.