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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 \( f \).

Ask by Mccoy Morrison. in the United States
Mar 09,2025

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Answer

The function \( f \) is not differentiable at \( x=1 \).

Solution

**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.

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To determine whether \( f \) is differentiable at \( x = 1 \), we first need to check if it is continuous at that point. The function is defined as \( f(x) = x^2 + 3 \) when \( x < 1 \) and \( f(x) = x + 3 \) when \( x \geq 1 \). Calculating the limit from the left as \( x \) approaches 1: \[ \lim_{x \to 1^-} f(x) = 1^2 + 3 = 4 \] Calculating the limit from the right as \( x \) approaches 1: \[ \lim_{x \to 1^+} f(x) = 1 + 3 = 4 \] Since both one-sided limits equal \( 4 \), \( f(1) = 4 \) ensures continuity at that point. Next, we examine differentiability by checking the derivatives from the left and right: \[ f'(x) = \begin{cases} 2x & \text{if } x < 1 \\ 1 & \text{if } x \geq 1 \end{cases} \] Calculating the left-hand derivative at \( x = 1 \): \[ \lim_{x \to 1^-} f'(x) = 2(1) = 2 \] Calculating the right-hand derivative at \( x = 1 \): \[ \lim_{x \to 1^+} f'(x) = 1 \] Since the left-hand derivative (2) does not equal the right-hand derivative (1), \( f \) is not differentiable at \( x = 1 \). No For the graph of \( f \), you'll see a parabola segment opening upwards starting at the point (1, 4) moving leftwards where the equation is \( f(x) = x^2 + 3 \). Right at that point, the graph transitions to a straight line with a slope of 1, starting at (1, 4) and moving upwards with the equation \( f(x) = x + 3 \). There will be a sharp corner at \( (1, 4) \), indicating it's continuous but not differentiable there.

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