Problem 11 - 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 \(x_{1}\). $$ f(x)=\sqrt[3]{x+1} ; x_{1}=-1 $$

Step-by-Step Solution

Answer

1a: Sketch 1b: Continuous 1c: Derivatives do not exist 1d: Not differentiable

Step 1 - Draw a sketch of the graph of the function

Start by sketching the graph of the function. The function is given by \[ f(x)=\root3{x+1} \] This is a cubic root function shifted to the left by 1 unit. Plot the function and label the key points, particularly the point \[ x_1 = -1 \] where the function value is \[ f(-1) = \root3{-1+1} = \root3{0} = 0 \]

Step 2 - Determine if f is continuous at x_1

To check for continuity at \[ x_1 = -1 \] , evaluate the following three conditions:1. \[ f(-1) = 0 \] 2. The limit from the left: \[ \lim_{x\to-1^-} \root3{x+1} = 0 \] 3. The limit from the right: \[ \lim_{x \to -1^+} \root3{x+1} = 0 \] Since all three conditions are equal, \[ f \] is continuous at \[ x_1 = -1 \]

Step 3 - Find f'(-1) if it exists

Differentiate the function: \[ f(x) = \root3{x+1} \] . Using the chain rule: \[ f'(x) = \frac{d}{dx}[x+1] \frac{1}{3} (x+1)^{-2/3} \] which simplifies to: \[ f'(x) = \frac{1}{3}(x+1)^{-2/3} \] . Evaluate the derivative at \[ x = -1 \]: \[ f'(-1) = \frac{1}{3}(-1+1)^{-2/3} = \frac{1}{3}(0) = \text{undefined} \]

Step 4 - Determine if f'_+(-1) exists

Check the right-hand derivative, for this, use the limit definition: \[ f_+'(-1) = \lim_{h \to 0^+} \frac{f(-1 + h) - f(-1)}{h} \lim_{h \to 0^+} \frac{ \root3{-1+h+1} - 0}{h} = \lim_{h \to 0^+} \frac{ \root3{h}}{h} = \lim_{h \to 0^+} \frac{h^{1/3}}{h} = \lim_{h \to 0^+} h^{-2/3} \] Since \[ h^{-2/3} \to \mathrm{\frac{1}{0}= \text{undefined}} \]

Step 5 - Determine Differentiability at x_1

Since the function's derivative does not exist at \[ x_1 = -1 \], \[ f \] is not differentiable at \[ x_1 = -1 \]

Key Concepts

Continuity of Functions

A function is continuous at a point if there is no interruption in the graph at that point. For a function to be continuous at a point \ \(x_1\ \), three criteria must be met:

• The function's value at \ \(x_1\ \) should exist:\ \(f(x_1)\ \)
• The limit of the function as \ \(x\ \) approaches \ \(x_1\ \) from the left should match the limit from the right:\ \( \lim_{ x \to x_1^-} f(x) \ = \lim_{x \to x_1^+} f(x)\ \)
• The value from the second condition should be equal to the function's value at that point: \ \( \lim_{ x \to x_1} f(x) \ = f(x_1)\ \)

Applying these to the function \ \(f(x)=\root3{ x+1}\ \) at \ \( x_1 = -1\ \), we see:
\ \( f(-1) = 0 \ \), \ \(\lim_{ x \to -1^-} \root3{x+1} = 0 \ \) and \ \( \lim_{ x \to -1^+} \root3{x+1} = 0 \ \). Since all these equal \ \(0\ \), the function is continuous at \ \(x_1 = -1\ \).

Limit Calculation

Calculating the limit of a function involves finding the value the function approaches as the input approaches a specific number. To find the limit of \ \(f(x) = \root3{x+1}\ \) at \ \( x = -1\ \), you consider two limits:

• Left-hand limit (as \ \( x\ \) approaches -1 from the negative side): \ \( \lim_{ x \to -1^-} \root3{x+1} = 0 \ \)
• Right-hand limit (as \ \( x\ \) approaches -1 from the positive side): \ \( \lim_{ x \to -1^+} \root3{x+1} = 0 \ \)

Since both these limits are equal and match the function's value at that point, we can conclude that \ \( \lim_{ x \to -1} \root3{x+1} = 0 \ \). It confirms continuity and equal limits from both sides.

Derivative Calculation

The derivative of a function at a point measures the rate at which the function's value changes as the input changes. To find the derivative of \ \( f(x) = \root3{ x+1}\ \), we use the chain rule:
\ \( f'(x) = \frac{d}{dx} [ x+1 ] \times \frac{1}{3} ( x+1 )^{-2/3} = \frac{1}{3} ( x+1 )^{-2/3} \ \).
At \ \(x = -1\ \), this form yields: \ \( f'(-1) = \frac{1}{3} (-1+1)^{-2/3} = \frac{1}{3} (0)^{-2/3} = undefined \ \). Since zero's power negative yields an undefined term, \ \(f'(-1)\ \) does not exist. Calculating the right-hand derivative using the limit definition confirms this:
\ \( \ f_+'(-1) = \lim_{h \to 0^+} \frac{ \root3{-1+h+1 } {- 0 } {/h} => \lim_{h \to 0^+} \frac{ h^{ 1/3}}{ h} = \lim_{h \to 0^+} h^{- 2/3} \ \ = \frac{ 1}{0 \ \), yields undefined results too.

Graph Sketching

Graph sketching helps visualize the behavior of a function. For \ \(f(x) = \root3{ x+1 }\ \), it's a cubic root function shifted left by one unit. Key points and behaviors to note:

• At \ \( x = -1\ \), the function crosses the x-axis at \ \( f(-1) = 0 \ \).
• For \ \( x > -1\ \), the function values positive but approaching zero more gradually as \ \( x-> -1\ \) from the right.
• For \ \( x < -1\ \), function values negative, again approaching zero more gradually.
Sketching key behavior helps understand continuity and differentiability, clearly showing no sharp turns or undefined jumps but rather consistent, smooth transitions.