Problem 7
Question
Do each of the following: (a) Draw a sketch of the graph of the function; (b) determine if \(f\) is continuous at \(x_{1} ;\) (c) find \(f^{\prime}-\left(x_{1}\right)\) and \(f_{+}^{\prime}\left(x_{1}\right)\) if they exist; (d) determine if \(f\) is differentiable at \(x_{1}\). $$ \begin{gathered} f(x)= \begin{cases}x^{2} & \text { if } x \leq 0 \\ -x^{2} & \text { if } x>0\end{cases} \\ x_{1}=0 \end{gathered} $$
Step-by-Step Solution
Verified Answer
The function is continuous and differentiable at \(x_1 = 0\). The left-hand and right-hand derivatives are both 0.
1Step 1: Sketch the Function
To draw the graph of the function, consider the different expressions for different intervals. For \(x \leq 0\), the function is \(f(x) = x^2\), which is a parabola opening upwards. For \(x > 0\), the function is \(f(x) = -x^2\), which is a parabola opening downwards. At \(x = 0\), both parabolas meet at the origin (0,0).
2Step 2: Determine Continuity at \(x_1 = 0\)
A function \(f(x)\) is continuous at \(x_1\) if \(\text{lim}_{x \to x_1^{-}} f(x) = \text{lim}_{x \to x_1^{+}} f(x) = f(x_1)\). Evaluate the left-hand limit and right-hand limit at \(x = 0\). \( \text{lim}_{x \to 0^{-}} f(x) = \text{lim}_{x \to 0^{-}} x^2 = 0 \) and \( \text{lim}_{x \to 0^{+}} f(x) = \text{lim}_{x \to 0^{+}} -x^2 = 0 \). Since both limits equal \(f(0) = 0\), \(f(x)\) is continuous at \(x = 0\).
3Step 3: Find the Derivative \(f'(x)\) and \(f'_+(x)\) at \(x_1 = 0\)
Calculate the derivatives from the left and right at \(x_1 = 0\). For \(x \leq 0\), \(f'(x) = \frac{d}{dx}(x^2) = 2x\). For \(x > 0\), \(f'_+(x) = \frac{d}{dx}(-x^2) = -2x\). Evaluate these at \(x_1 = 0\): \(f'(0) = 2 \times 0 = 0\) and \(f'_+(0) = -2 \times 0 = 0\).
4Step 4: Determine Differentiability at \(x_1 = 0\)
A function \(f(x)\) is differentiable at \(x_1\) if the left-hand and right-hand derivatives are equal. From Step 3, \(f'(0) = 0\) and \(f'_+(0) = 0\) are equal, thus \(f(x)\) is differentiable at \(x_1 = 0\).
Key Concepts
Continuity of FunctionsDerivative CalculationDifferentiability Analysis
Continuity of Functions
To understand the continuity of a function, we need to check if there are any breaks, jumps, or holes in the graph of the function. For a function to be continuous at a certain point, the following conditions must be met: The function’s value at that point must be defined. The left-hand limit as we approach the point must be the same as the right-hand limit. Both these limits should equal the function’s value at that point.
In our example, we are analyzing the function: \[ f(x) = \begin{cases} x^2 & \text{if } x \text{ ≤ 0}\ -x^2 & \text{if } x\text{ > 0} \end{cases} \] at \( x_1 = 0 \). To check for continuity at \( x_1 \), we calculate: \( lim_{x \to 0^-} f(x) \) and \( lim_{x \to 0^+} f(x) \). For \( x \text{ approaching from the left } \), we use \( x^2 \), which gives us \( 0 \). Similarly, approaching from the right using \( -x^2 \) also gives \( 0 \). Since both limits equal \( f(0) \) at \( 0 \), the function is continuous at \( x_1 = 0 \).
In our example, we are analyzing the function: \[ f(x) = \begin{cases} x^2 & \text{if } x \text{ ≤ 0}\ -x^2 & \text{if } x\text{ > 0} \end{cases} \] at \( x_1 = 0 \). To check for continuity at \( x_1 \), we calculate: \( lim_{x \to 0^-} f(x) \) and \( lim_{x \to 0^+} f(x) \). For \( x \text{ approaching from the left } \), we use \( x^2 \), which gives us \( 0 \). Similarly, approaching from the right using \( -x^2 \) also gives \( 0 \). Since both limits equal \( f(0) \) at \( 0 \), the function is continuous at \( x_1 = 0 \).
Derivative Calculation
Calculating the derivative of a function shows how it changes at any given point. Derivatives are the foundation of calculus, and they are used to find rates of change, slopes of curves, and optimize functions. To find the derivative of the given piecewise function:
\[ f(x) = \begin{cases} x^2 & \text{if } x\text{ ≤ 0}\ -x^2 & \text{if } x\text{ > 0} \end{cases} \] at \( x_1 = 0 \), we perform calculations separately for each piece. For \( x \text{ ≤ 0} \), the function is \( f(x) = x^2 \). Taking the derivative using differentiation rules, we get \( f'(x) = 2x \). For \( x > 0 \), where the function is \( f(x) = -x^2 \), the derivative is found to be \( f'_+(x) = -2x \). To find the derivative at \( x_1 =0 \), we evaluate these results at \( x=0 \): • \( f'(0) = 2 \times 0 = 0 \) • \( f'_+(0) = -2 \times 0 = 0 \) Therefore, both left-hand and right-hand derivatives have been calculated to be \( 0 \) at \( x_1 = 0 \).
\[ f(x) = \begin{cases} x^2 & \text{if } x\text{ ≤ 0}\ -x^2 & \text{if } x\text{ > 0} \end{cases} \] at \( x_1 = 0 \), we perform calculations separately for each piece. For \( x \text{ ≤ 0} \), the function is \( f(x) = x^2 \). Taking the derivative using differentiation rules, we get \( f'(x) = 2x \). For \( x > 0 \), where the function is \( f(x) = -x^2 \), the derivative is found to be \( f'_+(x) = -2x \). To find the derivative at \( x_1 =0 \), we evaluate these results at \( x=0 \): • \( f'(0) = 2 \times 0 = 0 \) • \( f'_+(0) = -2 \times 0 = 0 \) Therefore, both left-hand and right-hand derivatives have been calculated to be \( 0 \) at \( x_1 = 0 \).
Differentiability Analysis
A function is considered differentiable at a point if it is smooth and does not have any sharp corners or cusps at that point. In mathematical terms, a function is differentiable at a point if the left-hand and right-hand derivatives are equal at that point.
In our scenario, the function is given by: \[ f(x) = \begin{cases} x^2 & \text{if } x\text{ ≤ 0}\ -x^2 & \text{if } x\text{ > 0} \end{cases} \] with \( x_1 = 0 \). We previously found that both the left-hand derivative \( f'(0) \) and right-hand derivative \( f'_+(0) \) are equal to \( 0 \).
This means the function does not have any sharp changes and smoothly transitions through the point at \( x_1 = 0 \). Hence, we conclude that \( f(x) \) is differentiable at \( x_1=0 \). It's important to note that differentiability implies continuity, so because the function is differentiable at \( x_1 \), it is also confirmed to be continuous.
In our scenario, the function is given by: \[ f(x) = \begin{cases} x^2 & \text{if } x\text{ ≤ 0}\ -x^2 & \text{if } x\text{ > 0} \end{cases} \] with \( x_1 = 0 \). We previously found that both the left-hand derivative \( f'(0) \) and right-hand derivative \( f'_+(0) \) are equal to \( 0 \).
This means the function does not have any sharp changes and smoothly transitions through the point at \( x_1 = 0 \). Hence, we conclude that \( f(x) \) is differentiable at \( x_1=0 \). It's important to note that differentiability implies continuity, so because the function is differentiable at \( x_1 \), it is also confirmed to be continuous.
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