Consider the following. $$ f(x)=\left\{\begin{array}{ll}x^{2}+3 & \text { if } x

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Consider the following. $$ f(x)=\left\{\begin{array}{ll}x^{2}+3 & \text { if } x<1 \ x+3 & \text { if } x \geq 1\end{array}\right. $$ Is $$ f $$ differentiable at 1 ? Yes No Sketch the graph of $$ f $$.

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**Step 1. Check Continuity at $$ x=1 $$:**

- For $$ x<1 $$, the function is $$ f(x)=x^2+3 $$. So, the left-hand limit as $$ x $$ approaches 1 is
  $$
  \lim_{x\to 1^-} f(x) = 1^2 + 3 = 4.
  $$
- For $$ x\geq 1 $$, the function is $$ f(x)=x+3 $$. Therefore, the right-hand limit as $$ x $$ approaches 1 is
  $$
  \lim_{x\to 1^+} f(x) = 1 + 3 = 4.
  $$
- Since both one-sided limits are equal and $$ f(1)=4 $$, the function is continuous at $$ x=1 $$.

**Step 2. Check Differentiability at $$ x=1 $$:**

- For $$ x<1 $$, differentiate $$ f(x)=x^2+3 $$:
  $$
  f'_-(x) = 2x.
  $$
  Evaluating at $$ x=1 $$ gives
  $$
  f'_-(1) = 2\cdot1 = 2.
  $$

- For $$ x\geq 1 $$, differentiate $$ f(x)=x+3 $$:
  $$
  f'_+(x) = 1.
  $$
  Evaluating at $$ x=1 $$ gives
  $$
  f'_+(1) = 1.
  $$

- Since the left-hand derivative $$ 2 $$ is not equal to the right-hand derivative $$ 1 $$, the function is **not differentiable** at $$ x=1 $$.

**Answer to the First Question:**

No, $$ f $$ is not differentiable at 1.

**Step 3. Sketch the Graph of $$ f $$:**

- **For $$ x<1 $$:**
  - The graph is of the parabola $$ y=x^2+3 $$ which is a standard upward-opening parabola shifted upward by 3 units.
  - When $$ x $$ is close to 1 from the left, the point on the graph is $$ (1, 4) $$. Note that since the formula applies for $$ x < 1 $$, the point $$ (1,4) $$ is not actually part of this branch (represented as an open circle).

- **For $$ x\geq 1 $$:**
  - The graph is of the line $$ y=x+3 $$, which is a straight line with slope 1 and $$ y $$-intercept at 3.
  - At $$ x=1 $$, the line gives $$ y=1+3=4 $$, so the point $$ (1,4) $$ is included and is drawn as a solid dot.

- **Graph Description:**
  - Plot the parabola $$ y=x^2+3 $$ for $$ x < 1 $$. Near $$ x=1 $$, draw an open circle at $$ (1,4) $$.
  - For $$ x\geq 1 $$, draw the line $$ y=x+3 $$ starting at $$ (1,4) $$ with a solid dot.
  - This visual shows a continuous connection at $$ x=1 $$ but a "corner" where the left and right derivatives differ.

The final sketch will show a smooth parabola approaching the point $$ (1,4) $$ from the left and a straight line departing from $$ (1,4) $$ to the right, with the corner at $$ x=1 $$ indicating the lack of differentiability.
The function $$ f $$ is not differentiable at $$ x=1 $$.

Consider the following. \[ f(x)=\left\{\begin{array}{ll}x^{2}+3 & \text { if } x<1 \\ x+3 & \text { if } x \geq 1\end{array}\right. \] Is \( f \) differentiable at 1 ? Yes No Sketch the graph of \( f \).

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