Problem 12 - 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 \(x_{1}\). $$ f(x)=(x-2)^{-2} ; x_{1}=2 $$

Step-by-Step Solution

Answer

The function is not continuous and not differentiable at \(x = 2\) and its derivatives diverge.

Step 1 Title - Sketch the graph of the function

To sketch the graph of the function, first rewrite the function as a rational function: \[f(x) = \frac{1}{(x - 2)^2}\] This function has a vertical asymptote at \(x = 2\) because the denominator becomes zero there. The graph approaches infinity as \(x\) approaches \(2\) from either side and decreases to zero as \(x\) moves away from \(2\). Draw the graph with this information in mind.

Step 2 Title - Determine if the function is continuous at \(x = 2\)

A function is continuous at a point if the limit as \(x\) approaches the point is equal to the function value at that point. In this case, as \(x\) approaches 2, the function value approaches infinity: \[ \lim_{{x \to 2}} f(x) = \lim_{{x \to 2}} \frac{1}{(x - 2)^2} = \infty \] Since \(f(2)\) is not defined (the function tends towards infinity), \(f(x)\) is not continuous at \(x = 2\).

Step 3 Title - Find the left-hand and right-hand derivatives at \(x = 2\)

To find the derivatives, compute the left-hand derivative, \(f'_{-}(x_{1})\), and the right-hand derivative, \(f'_{+}(x_{1})\), using the limit definitions. \(f'_{-}(2)\): \[{f'_{-}(2) = \lim_{{h \to 0^-}} \frac {f(2 + h) - f(2)}{h}} = \lim_{{h \to 0^-}} \frac {\frac{1}{(2 + h - 2)^{2}} - \infty}{h}\] Since \(f(2)\) is not defined, this limit also diverges. \(f'_{+}(2)\): \[{f'_{+}(2) = \lim_{{h \to 0^+}} \frac {f(2 + h) - f(2)}{h}} = \lim_{{h \to 0^+}} \frac {\frac{1}{(2 + h - 2)^{2}} - \infty}{h}\] This limit also diverges because \(f(2)\) does not exist.

Step 4 Title - Determine if the function is differentiable at \(x = 2\)

For a function to be differentiable at a point, it must be continuous and the left-hand and right-hand derivatives must exist and be equal. From previous steps, it is already established that the function is not continuous at \(x = 2\) and the derivatives diverge. Hence, \(f(x)\) is not differentiable at \(x = 2\).

Key Concepts

Graph of Function

To understand the behavior of a function, start by sketching its graph. Consider the function given, \( f(x) = (x - 2)^{-2} \).
When rewritten, it becomes \( f(x) = \frac{1}{(x - 2)^2} \). This tells us several things about the behavior of the function:

• For \(xeq2\), the function is always positive and never zero, since squaring any non-zero number is always positive.
• As \(x\) approaches 2 from either side, the denominator of the function gets very small, causing the function value to become very large and approach infinity.
• Conversely, as \(x\) moves away from 2, the function value decreases and approaches zero.

By sketching these behaviors, you can see how the graph shows a vertical asymptote at \(x=2\) where the function approaches infinity. The curve will rise steeply, and as \(x\) moves far from 2, the function value will approach zero but never actually reach it.

Vertical Asymptote

A vertical asymptote is a line that the graph of a function approaches but never actually touches or crosses. For the function \(f(x) = \frac{1}{(x - 2)^2} \), there is a vertical asymptote at \(x = 2\).
Why? Because at \(x=2\), the denominator of the function becomes zero, which makes the function value undefined. In other words, the closer \(x\) gets to 2, the higher \(f(x)\) becomes.

To identify vertical asymptotes:

• Find the points where the denominator of a function is zero.
• Check what happens to the function value as \(x\) approaches those points.

For \(f(x) = \frac{1}{(x-2)^2}\), we find that as \(x\) approaches 2 from both sides (left and right), \(f(x)\) goes to infinity. Hence, \(x = 2\) is a vertical asymptote.

Limits and Continuity

Continuity at a point requires the function to be defined at that point and the limit of the function as it approaches that point should equal the function value there.
For the function \(f(x) = \frac{1}{(x-2)^2}\), let's check its continuity at \(x = 2\):

• The limit as \(x\) approaches 2 is \( \lim_{{x \to 2}} f(x) = \lim_{{x \to 2}} \frac{1}{(x - 2)^2} = \infty \)
• Since the function value at \(x = 2\) is not defined (it goes to infinity), \(f(x)\) is not continuous at \(x = 2\).

For a function to be continuous at a point:

• The function must be defined at the point.
• The limit of the function as it approaches the point must exist.
• The limit must equal the function value at that point.

Thus, \(f(x)\) fails all these conditions at \(x = 2\).

One-Sided Derivatives

One-sided derivatives analyze how a function behaves as it approaches a point from either the left or the right.
For the function \(f(x) = \frac{1}{(x - 2)^2}\) at \(x = 2\), we can compute the left-hand and right-hand derivatives.
Left-hand derivative (as \(h\) approaches zero from the left):ewline \(f'_{-}(2) = \lim_{h \to 0^-} \frac {f(2 + h) - f(2)}{h} = \lim_{h \to 0^-} \frac{\frac{1}{(2 + h - 2)^{2}} - \infty}{h}\)
Right-hand derivative (as \(h\) approaches zero from the right):ewline \(f'_{+}(2) = \lim_{h \to 0^+} \frac {f(2 + h) - f(2)}{h} = \lim_{h \to 0^+} \frac{\frac{1}{(2 + h - 2)^{2}} - \infty}{h}\)
Because \(f(2)\) is not defined and both limits approach infinity, neither the left-hand nor the right-hand derivative exists at \(x=2\).
For a function to be differentiable at a point, it must be continuous at that point and the left-hand and right-hand derivatives at that point must both exist and be equal. Given the divergence at \(x=2\), \(f(x)\) is not differentiable at this point.