Given the function f(x) = cases 1/(1+x²), x < dfrac12

Given the function $f(x) = \begin{cases} \dfrac{1}{1+x^{2}}, & x\lt \dfrac12,\\[8pt] \dfrac25x+\dfrac35, & \dfrac12\lt x\lt 1,\\[8pt] x^{2}-6x+6, & x\gt 1, \end{cases}$ investigate 1° its continuity; 2° the derivative and its continuity; 3° the extrema, and the least and greatest values on $\left[\dfrac35,\dfrac52\right]$; 4° sketch the graph. (The function is not defined at $x = \tfrac12$ and $x = 1$.)

Step 1: 1° — continuity at $x = \tfrac12$

\[ \lim_{x\to(1/2)^{-}}f(x) = \frac{1}{1+\tfrac14} = \frac{4}{5}, \qquad \lim_{x\to(1/2)^{+}}f(x) = \frac25\cdot\frac12+\frac35 = \frac15+\frac35 = \frac45. \]

The one-sided limits agree, so $\displaystyle\lim_{x\to1/2}f(x) = \frac45$ exists. As $f$ is not defined there, the discontinuity is removable: setting $f\!\left(\tfrac12\right) = \tfrac45$ makes $f$ continuous.

Step 2: 1° — continuity at $x = 1$

\[ \lim_{x\to1^{-}}f(x) = \frac25+\frac35 = 1, \qquad \lim_{x\to1^{+}}f(x) = 1-6+6 = 1. \]

Again the limits agree and the discontinuity is removable, with $f(1) = 1$. Elsewhere each formula is continuous, so $f$ is continuous on $\mathbb{R}\setminus\left\{\tfrac12,1\right\}$ and extends continuously to all of $\mathbb{R}$.

Step 3: 2° — the derivative

\[ f'(x) = \begin{cases} \dfrac{-2x}{\bigl(1+x^{2}\bigr)^{2}}, & x\lt \dfrac12,\\[8pt] \dfrac25, & \dfrac12\lt x\lt 1,\\[8pt] 2x-6, & x\gt 1. \end{cases} \]

Step 4: 2° — the derivative is discontinuous at both joins

\[ \lim_{x\to(1/2)^{-}}f'(x) = \frac{-1}{\left(\tfrac54\right)^{2}} = -\frac{16}{25}, \qquad \lim_{x\to(1/2)^{+}}f'(x) = \frac25, \]

\[ \lim_{x\to1^{-}}f'(x) = \frac25, \qquad \lim_{x\to1^{+}}f'(x) = 2-6 = -4. \]

Since $-\dfrac{16}{25}\neq\dfrac25$ and $\dfrac25\neq-4$, the derivative has a jump discontinuity at each join. Even after the removable discontinuities of $f$ are repaired, the extended function has corners there and is not differentiable at $x = \tfrac12$ or $x = 1$.

Step 5: 3° — extrema

Interior zeros of $f'$. On the first piece $f'(x) = 0$ at $x = 0$, where $f'$ passes from $+$ to $-$: a local maximum, $f(0) = 1$. On the second piece $f' = \tfrac25\gt 0$ throughout — no critical point. On the third piece $f'(x) = 2x-6 = 0$ at $x = 3$, where $f'$ passes from $-$ to $+$: a local minimum, $f(3) = 9-18+6 = -3$.

The joins. A corner can also be an extremum, and here both are. At $x = \tfrac12$ the slope jumps from $-\tfrac{16}{25}$ to $+\tfrac25$, a change from $-$ to $+$: a local minimum of value $\tfrac45$. At $x = 1$ it jumps from $+\tfrac25$ to $-4$, a change from $+$ to $-$: a local maximum of value $1$.

\[ \begin{array}{c|c|c} x & \text{type} & f(x) \\ \hline 0 & \text{local maximum} & 1 \\ \tfrac12 & \text{local minimum (corner)} & \tfrac45 \\ 1 & \text{local maximum (corner)} & 1 \\ 3 & \text{local minimum} & -3 \end{array} \]

Step 6: 3° — least and greatest values on $\left[\tfrac35,\tfrac52\right]$

The interval meets the second and third pieces. On $\left[\tfrac35,1\right)$ the function increases with slope $\tfrac25$; on $\bigl(1,\tfrac52\bigr]$ it decreases, since $f' = 2x-6\lt 0$ for $x\lt 3$ and the critical point $x = 3$ lies outside. Hence the candidates are the two endpoints and the join:

\[ f\!\left(\frac35\right) = \frac25\cdot\frac35+\frac35 = \frac{21}{25} = 0.84, \qquad f(1) = 1, \qquad f\!\left(\frac52\right) = \frac{25}{4}-15+6 = -\frac{11}{4}. \]

\[ \max_{\left[3/5,\,5/2\right]}f = f(1) = 1, \qquad \min_{\left[3/5,\,5/2\right]}f = f\!\left(\frac52\right) = -\frac{11}{4}. \]

The maximum is attained only because the removable discontinuity at $x = 1$ has been repaired by setting $f(1) = 1$; for the original function, $1$ is a supremum that is approached but never reached.

Step 7: 4° — the graph

Purple: $\dfrac{1}{1+x^{2}}$; red: $\dfrac25x+\dfrac35$; green: $x^{2}-6x+6$. The open circles mark the two removable joins, where the value is missing but the limit exists.

Step 8: Verification

\[f(0.4) = 0.8620690 \gt f(0.5^{\pm}) = 0.8 \lt f(0.6) = 0.84 \;\;\checkmark\ \text{(a genuine corner minimum)}\]

\[f(0.9) = 0.96 \lt f(1) = 1 \gt f(1.2) = 0.24 \;\;\checkmark\ \text{(a corner maximum)}\]

\[f(2.5) = -2.75 \gt f(3) = -3 \;\;\checkmark\ \text{(the minimum lies beyond the interval)}\]