Problem 9 - 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 \(x_{1}\). $$ \begin{gathered} f(x)= \begin{cases}\sqrt{1-x} & \text { if } x<1 \\ (1-x)^{2} & \text { if } x \geq 1\end{cases} \\ & x_{1}=1 \end{gathered} $$

Step-by-Step Solution

Answer

The function \( f \) is continuous at \( x_{1} = 1 \) but not differentiable at \( x_{1} = 1 \).

Step 1: Draw the sketch of the graph

To draw the graph of the function, break it into two pieces based on the given conditions: For \( x < 1 \), the function is \( f(x) = \sqrt{1 - x} \), which is a decreasing curve that starts from (0, 1) and approaches (1, 0). For \( x \geq 1 \), the function is \( f(x) = (1 - x)^{2} \), which is a parabola opening upwards starting from (1, 0). Plot these pieces separately and note that they meet at (1, 0).

Step 2: Determine if \( f \) is continuous at \( x_{1} \)

Check the value of the function at \( x = 1 \). For continuity at \( x_{1} = 1 \), \( \lim\limits_{x \to 1^{-}} f(x) \) must equal \( \lim\limits_{x \to 1^{+}} f(x) \). Calculate: \( \lim \limits_{x \to 1^{-}} f(x) = \lim \limits_{x \to 1^{-}} \sqrt{1 - x} = \sqrt{0} = 0 \) \( f(1) = (1 - 1)^{2} = 0 \) Thus, \( f \) is continuous at \( x_{1} = 1 \).

Step 3: Find \( f_{-}^{\prime}(1) \) and \( f_{+}^{\prime}(1) \)

Calculate the left-hand and right-hand derivatives at \( x = 1 \). For \( f_{-}^{\prime}(1) \): \( f(x) = \sqrt{1 - x} \) Use the definition of the derivative, \( f_{-}^{\prime}(1) = \lim\limits_{h \to 0^{-}} \( \frac{\sqrt{1 - (1 + h)} - \sqrt{1 - 1}}{h} \) = \lim\limits_{h \to 0^{-}} \( \frac{\sqrt{-h}}{h} \) = \lim\limits_{h \to 0^{-}} \( -\frac{1}{2\sqrt{1 - x}} \) \biggr |_{x=1} = -\infty \)For \( f_{+}^{\prime}(1) \): \( f(x) = (1 - x)^{2} \) Use the definition of the derivative, \( f_{+}^{\prime}(1) = \lim\limits_{h \to 0^{+}} \( \frac{(1 - (1 + h))^{2} - 0}{h} \) = \lim\limits_{h \to 0^{+}} \( \frac{h^{2}}{h} \) = \lim\limits_{h \to 0^{+}} h = 0 \)

Step 4: Determine if \( f \) is differentiable at \( x_{1} \)

The function \( f \) is differentiable at \( x = 1 \) if both the limit of the left-hand derivative and right-hand derivative exist and are equal. We already found: \( f_{-}^{\prime}(1) = -\infty \) and \( f_{+}^{\prime}(1) = 0 \), Since these are not equal, \( f \) is not differentiable at \( x_{1} = 1 \).

Key Concepts

graphing piecewise functions

A piecewise function is defined by different expressions depending on the value of the independent variable. In this exercise, we have a function defined as follows:
For \( x < 1 \), the function is \( f(x) = \sqrt{1 - x} \). This part of the function is a decreasing curve, starting from the point (0, 1) and moving towards (1, 0).
For \( x \geq 1 \), the function is \( (1 - x)^2 \). This defines a parabola that starts at the point (1, 0) and opens upwards.
To graph the function, plot the points and curves separately for each piece and note that they meet at (1, 0). It is important to clearly mark the transition point where the behavior of the function changes.

continuity at a point

Continuity at a point means that there are no breaks, jumps, or holes at that point in the graph of the function. For a function \( f \) to be continuous at a point \( x_1 \), the following conditions must be met:

• \( \lim_{x \to x_1^-} f(x) \) exists
• \( \lim_{x \to x_1^+} f(x) \) exists
• These two limits are equal to each other and to \( f(x_1) \)

In the given exercise, the function transition point is at \( x_1 = 1 \). We calculate:
\( \lim_{x \to 1^-} f(x) = \sqrt{1 - 1} = 0 \)
\( \lim_{x \to 1^+} f(x) = (1 - 1)^2 = 0 \)
\( f(1) = (1 - 1)^2 = 0 \)
Since all the above values are equal, we conclude that \( f \) is continuous at \( x_1 = 1 \).

left-hand and right-hand derivatives

To determine the differentiability at a point, it is crucial to understand the concept of left-hand and right-hand derivatives. The left-hand derivative at a point \( x_1 \), denoted \( f'_- (x_1) \), is the derivative as we approach the point from the left (values less than \( x_1 \)). The right-hand derivative, denoted \( f'_+ (x_1) \), is the derivative as we approach the point from the right (values greater than or equal to \( x_1 \)).

In the exercise, we compute:
For \( x < 1 \), \( f(x) = \sqrt{1 - x} \)
\( f'_- (1) = \lim_{h \to 0^-} \frac{\sqrt{1 - (1+h)} - \sqrt{1 - 1}}{h} = \lim_{h \to 0^-} \frac{\sqrt{-h}}{h} = \lim_{h \to 0^-} -\frac{1}{2\sqrt{h}}|_{x=1} = -\infty \)
For \( x \geq 1 \), \( f(x) = (1 - x)^2 \)
\( f'_+ (1) = \lim_{h \to 0^+} \frac{(1 - (1+h))^2 - 0}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0 \)
So, \( f'_- (1) = -\infty \) and \( f'_+ (1) = 0 \).

differentiability criteria

A function is differentiable at a point if its graph has a tangent line that is well-defined at that point. This requires that both the left-hand and right-hand derivatives exist and are equal.
From the exercise, we know:
\( f'_- (1) = -\infty \)
\( f'_+ (1) = 0 \)
Since the left-hand derivative and the right-hand derivative are not equal, \( f \) is not differentiable at \( x_1 = 1 \). This means the graph has a sharp corner or cusp at this point, preventing a single tangent line from being drawn.