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a. Show that the definition of the derivative applied to the function \(f(x)=\sqrt{x}\) at \(x=0\) gives \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\sqrt{h}}{h}\). b. Use a calculator to evaluate the difference quotient \(\frac{\sqrt{h}}{h}\) for the following values of \(h\) : \(0.1,0.001\), and \(0.00001\). [Hint: Enter the calculation into your calculator with \(h\) replaced by \(0.1\), and then change the value of \(h\) by inserting zeros.] c. From your answers to part (b), does the limit exist? Does the derivative of \(f(x)=\sqrt{x}\) at \(x=0\) exist? d. Graph \(f(x)=\sqrt{x}\) on the window \([0,1]\) by \([0,1]\). Do you see why the slope at \(x=0\) does not exist?

Short Answer

Expert verified
The derivative \( f'(0) \) does not exist due to a non-finite limit and a cusp at \( x = 0 \).

Step by step solution

01

Understand the problem

We are given the function \( f(x) = \sqrt{x} \) and need to find its derivative at \( x = 0 \) using the limit definition of a derivative. Then, we evaluate this derivative using specific values of \( h \). Finally, we visualize the graph to determine whether the derivative exists at \( x = 0 \).
02

Apply Limit Definition of Derivative

The derivative of \( f(x) \) at \( x = 0 \) is given by \( f'(0) = \lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h} \). Substituting \( f(x) = \sqrt{x} \), we have \( f'(0) = \lim_{h \rightarrow 0} \frac{\sqrt{h} - \sqrt{0}}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{h}}{h} \). Thus, \( f'(0) = \lim_{h \rightarrow 0} \frac{\sqrt{h}}{h} \).
03

Evaluate the Difference Quotient

Calculate \( \frac{\sqrt{h}}{h} \) for \( h = 0.1, 0.001, \text{ and } 0.00001 \). Using a calculator:- For \( h = 0.1 \), \( \frac{\sqrt{0.1}}{0.1} \approx 3.162 \).- For \( h = 0.001 \), \( \frac{\sqrt{0.001}}{0.001} \approx 31.622 \).- For \( h = 0.00001 \), \( \frac{\sqrt{0.00001}}{0.00001} \approx 316.2278 \).
04

Analyze Limit and Derivative Existence

Observing the values from Step 3, as \( h \) becomes smaller, \( \frac{\sqrt{h}}{h} \) increases significantly. This suggests that the limit does not converge to a finite number, indicating that the derivative of \( f(x) = \sqrt{x} \) at \( x = 0 \) does not exist.
05

Graph the Function

Graph \( f(x) = \sqrt{x} \) on the window \([0, 1] \times [0, 1]\). At \( x = 0 \), the graph shows a sharp corner or cusp, emphasizing a sudden change in direction. This visualizes why the slope at \( x = 0 \) does not exist, as the graph doesn't smoothly pass through that point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Definition of Derivative
The limit definition of a derivative is a fundamental concept in calculus. It allows us to determine the instantaneous rate of change of a function at any given point. For any function \( f(x) \), the derivative at a particular point \( x = a \) is given by the limit: \[ f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} \] This formula essentially computes the slope of the tangent line to the curve \( f(x) \) at \( x = a \). In the original exercise, we applied this definition to \( f(x) = \sqrt{x} \) at \( x = 0 \). By substituting, we found: \[ f'(0) = \lim_{h \rightarrow 0} \frac{\sqrt{h}}{h} \] This expression requires us to examine the behavior of \( \frac{\sqrt{h}}{h} \) as \( h \) approaches zero to decide on the existence of the derivative. Using actual numbers for \( h \) allows us to approximate this limit, but as seen with decreasing \( h \), the result keeps increasing, hinting at some underlying math issue.
Non-Existence of Derivative
A derivative might not exist at a certain point like in our function example \( f(x)=\sqrt{x} \) at \( x=0 \). Let’s explore why. There are several reasons a derivative may not exist at a point:
  • The function might be discontinuous at that point.
  • The function may have a sharp corner or cusp.
  • There might be a vertical tangent line with infinite slope.
For \( f(x)=\sqrt{x} \), it's the cusp scenario. This arises because the expression \( \lim_{h \rightarrow 0} \frac{\sqrt{h}}{h} \) does not settle down to a finite value. As we calculated: - For \( h = 0.1 \), the result was approximately 3.162. - For smaller \( h \) values \( 0.001 \) and \( 0.00001 \), the results escalated to 31.622 and 316.228, respectively. This indicates the slope approaches infinity as it stretches to smaller \( h \). Therefore, the derivative—meaning the slope of the tangent—does not exist at that point.
Graphical Analysis of Functions
Graphing a function is a powerful way to visualize concepts in calculus, such as derivatives. The function \( f(x) = \sqrt{x} \) presents something particularly important when analyzed visually on a graph. In the graph of \( f(x) = \sqrt{x} \) over the window \([0, 1]\) by \([0, 1]\), you can spot why the derivative at \( x = 0 \) doesn't exist. The graph shows a distinctive cusp at \( x = 0 \). Here’s what to look for:
  • At \( x = 0 \), instead of a smooth curve or gentle turning point, you see a sharp corner.
  • Such sharp corners or cusps suggest infinite and undefined slopes.
  • This graphical behavior aligns perfectly with the algebraic concept we explored, where the slope doesn't stabilize.
Thus, an important lesson is learned: graphical analysis can be used to confirm the characteristics suggested by the calculus of derivatives. Through visuals, we better grasp why at a cusp, a derivative is non-existent.

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