Problem 2 - The Calculus with Analytic Geometry

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 \(x_{1}\). $$ f(x)=\left\\{\begin{array}{cc} 3-2 x & \text { if } x<2 \\ 3 x-7 & \text { if } x \geq 2 \\ & x_{1}=2 \end{array}\right. $$

Step-by-Step Solution

Answer

The function is continuous at \(x_{1} = 2\), but it is not differentiable at \(x_{1} = 2\). The left-hand derivative is \( -2\) and the right-hand derivative is \( 3\).

Step 1: Draw the graph of the function

The function is piecewise, so consider the two cases separately. For \(x < 2\), the function is \(f(x) = 3 - 2x\). For \(x \geq 2\), the function is \(f(x) = 3x - 7\). Plot these two segments on the coordinate plane.

Step 2: Check continuity at \(x = 2\)

To check if \(f\) is continuous at \(x_{1} = 2\), evaluate the left-hand limit (LHL) and the right-hand limit (RHL) as \(x\) approaches 2, and compare them with \(f(2)\).\[ \text{LHL:} \lim_{{x \to 2^-}} (3 - 2x) = 3 - 4 = -1 \] \[ \text{RHL:} \lim_{{x \to 2^+}} (3x - 7) = 6 - 7 = -1 \] The limits match and \(f(2) = -1\). Thus, \(f\) is continuous at \(x_{1} = 2\).

Step 3: Find the left-hand derivative \(f'_{-}(2)\) if it exists

Using the definition of the derivative, calculate the left-hand derivative at \(x_{1} = 2\).\[ f'_{-}(2) = \lim_{{h \to 0^-}} \frac{{f(2 + h) - f(2)}}{h} \] Since \(f(x) = 3 - 2x\) for \(x < 2\),\[ f'_{-}(2) = \lim_{{h \to 0^-}} \frac{{(3 - 2(2 + h)) - (-1)}}{h} = \lim_{{h \to 0^-}} \frac{{-2h}}{h} = -2 \]

Step 4: Find the right-hand derivative \(f'_{+}(2)\) if it exists

Using the definition of the derivative, calculate the right-hand derivative at \(x_{1} = 2\).\[ f'_{+}(2) = \lim_{{h \to 0^+}} \frac{{f(2 + h) - f(2)}}{h} \] Since \(f(x) = 3x - 7\) for \(x \geq 2\),\[ f'_{+}(2) = \lim_{{h \to 0^+}} \frac{{(3(2 + h) - 7) - (-1)}}{h} = \lim_{{h \to 0^+}} \frac{{3h}}{h} = 3 \]

Step 5: Determine if \(f\) is differentiable at \(x_{1} = 2\)

The function \(f\) is differentiable at \(x_{1}\) if both left-hand and right-hand derivatives exist and are equal.From Steps 3 and 4, \(f'_{-}(2) = -2\) and \(f'_{+}(2) = 3\). Since these are not equal, \(f\) is not differentiable at \(x_{1} = 2\).

Key Concepts

Understanding Continuity

Continuity is an important concept in calculus that describes how a function behaves at a specific point and around it. A function is continuous at a point if there are no breaks, jumps, or holes at that point. For a function to be continuous at a point \(x = a\), three conditions must be satisfied:

• The function \(f(x)\) is defined at \(a\), meaning \(f(a)\) exists.
• The limit of the function as \(x\) approaches \(a\) exists. This means both the left-hand limit (LHL) and the right-hand limit (RHL) as \(x\) approaches \(a\) must exist.
• The limit of the function as \(x\) approaches \(a\) is equal to the value of the function at \(a\), i.e., \( \lim_{{x \to a}} f(x) = f(a) \).

For the given piecewise function, we checked for continuity at \(x = 2\). We calculated the LHL and RHL and found they both equal to -1. Since \(f(2)= -1\), the function is continuous at \(x = 2\).

Exploring Differentiability

Differentiability is a measure of how smooth a function is. If a function is differentiable at a point, it means the function has a well-defined tangent at that point, and therefore, a unique slope. For a function \(f\) to be differentiable at a point \(x = a\), the following must hold:

• The function \(f\) must be continuous at \(a\). Continuity is a prerequisite for differentiability.
• Both the left-hand derivative \(f'_{-}(a)\) and the right-hand derivative \(f'_{+}(a)\) exist.
• The left-hand and right-hand derivatives must be equal, i.e., \(f'_{-}(a) = f'_{+}(a)\).

In our problem, we calculated the left-hand derivative at \(x = 2\), which was \( -2 \), and the right-hand derivative at \(x = 2\), which was \( 3 \). Since these derivatives are not equal, the function is not differentiable at \(x = 2\). Therefore, while the function is continuous at \(x = 2\), it is not smooth enough at that point to have a well-defined tangent or slope.

Calculating Derivatives

A derivative represents the rate of change of a function with respect to a variable. It can be seen as the slope of the tangent line to the function's graph at a particular point. The derivative of a function \(f(x)\) at a point \(x = a\) is given by the limit: \[ f'(a) = \lim_{{h \to 0}} \frac{{f(a + h) - f(a)}}{h} \] For piecewise functions, we often need to find the left-hand and right-hand derivatives separately:

• The left-hand derivative (LHD) at \(x = a\) is given by: \[ f'_{-}(a) = \lim_{{h \to 0^-}} \frac{{f(a + h) - f(a)}}{h} \]
• The right-hand derivative (RHD) at \(x = a\) is given by: \[ f'_{+}(a) = \lim_{{h \to 0^+}} \frac{{f(a + h) - f(a)}}{h} \]

In our given piecewise function: \
For \(x < 2\): \( f(x) = 3 - 2x \)
For \(x \geq 2\): \( f(x) = 3x - 7 \)
To find \(f'_{-}(2)\), we used the definition for the left segment and calculated: \[ f'_{-}(2) = \lim_{{h \to 0^-}} \frac{{(3 - 2(2 + h)) - (-1)}}{h} = -2 \]
For \(f'_{+}(2)\), we used the definition for the right segment and calculated: \[ f'_{+}(2) = \lim_{{h \to 0^+}} \frac{{(3(2 + h) - 7) - (-1)}}{h} = 3 \] Thus, the derivatives on either side of \(x = 2\) are not equal, confirming that \(f\) is not differentiable at that point.