Question

A function \(f\) and \(a\) value \(a\) are given. Approximate the limit of the difference quotient, \(\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h},\) using \(h=\pm 0.1, \pm 0.01 .\) $$ f(x)=\frac{1}{x+1}, \quad a=2 $$

Short answer

The derivative of the function at \(a = 2\) is approximately \(-0.1\).

Step-by-step solution

Calculate Function Values

Calculate \(f(2 + h)\) and \(f(2)\). Using the function \(f(x) = \frac{1}{x+1}\), \[f(2) = \frac{1}{2+1} = \frac{1}{3}\]. For the values of \(h\), calculate: - \(f(2 + 0.1) = \frac{1}{(2 + 0.1) + 1} = \frac{1}{3.1}\) - \(f(2 - 0.1) = \frac{1}{(2 - 0.1) + 1} = \frac{1}{2.9}\) - \(f(2 + 0.01) = \frac{1}{(2 + 0.01) + 1} = \frac{1}{3.01}\) - \(f(2 - 0.01) = \frac{1}{(2 - 0.01) + 1} = \frac{1}{2.99}\).

Calculate Difference Quotients

Using the function values from Step 1, calculate the difference quotient for each chosen value of \(h\): - For \(h = 0.1\): - \(\frac{f(2 + 0.1) - f(2)}{0.1} = \frac{\frac{1}{3.1} - \frac{1}{3}}{0.1}\) - \(\frac{f(2 - 0.1) - f(2)}{-0.1} = \frac{\frac{1}{2.9} - \frac{1}{3}}{-0.1}\)- For \(h = 0.01\): - \(\frac{f(2 + 0.01) - f(2)}{0.01} = \frac{\frac{1}{3.01} - \frac{1}{3}}{0.01}\) - \(\frac{f(2 - 0.01) - f(2)}{-0.01} = \frac{\frac{1}{2.99} - \frac{1}{3}}{-0.01}\).

Simplify and Evaluate Approximations

Now, simplify the expressions from Step 2:- For \(h = 0.1\): - \(\frac{f(2 + 0.1) - f(2)}{0.1} = \frac{\frac{1}{3.1} - \frac{1}{3}}{0.1} \approx -0.1075\) - \(\frac{f(2 - 0.1) - f(2)}{-0.1} = \frac{\frac{1}{2.9} - \frac{1}{3}}{-0.1} \approx -0.1034\)- For \(h = 0.01\): - \(\frac{f(2 + 0.01) - f(2)}{0.01} = \frac{\frac{1}{3.01} - \frac{1}{3}}{0.01} \approx -0.0997\) - \(\frac{f(2 - 0.01) - f(2)}{-0.01} = \frac{\frac{1}{2.99} - \frac{1}{3}}{-0.01} \approx -0.1003\).

Conclusion

The average of the approximations as \(h\) approaches 0 gives the best estimate: Averaging the results for very small \(h\) provides a more accurate estimate of the derivative. For \(h = 0.01\), average of \(-0.0997\) and \(-0.1003\) is approximately \(-0.1\).Thus, the limit of the difference quotient, which is the derivative \(f'(2)\), is approximately \(-0.1\).

Key concepts

Limit of a Function

The concept of the limit of a function is fundamental in calculus. It refers to the value that a function approaches as the input, or variable, gets closer to a certain point. When we talk about \(\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h},\)we are interested in what happens to the expression as \(h\) becomes extremely small.
This specific form is known as the difference quotient. As \(h\) tends towards zero, the quotient tells us how much the function \(f(x)\) is changing at \(a\).
Understanding limits is crucial as it lays the groundwork for defining derivatives and continuity. Limits allow us to deal with behaviors at points where direct substitution might not work.

Derivative

The derivative of a function gives us the rate at which the function value changes with respect to changes in the input, often referred to as the slope of the function at a particular point. In simple terms, the derivative at a point provides the instantaneous rate of change.
Using the limit of the difference quotient as \(h \rightarrow 0\), which you've calculated as approximately \(-0.1\) for \(f'(2)\), tells us how steep the function is at \(x=2\). This negative value indicates the function is decreasing at this point.

• Finding derivatives is akin to zooming in on the graph of the function until it looks like a straight line, allowing you to determine the slope of that line.
• Derivatives are used across calculus to solve problems involving rates of change and motion.

Approximation

Approximating values is a valuable technique, especially in calculus, where exact values are difficult to calculate.
In the exercise, using different small values of \(h = \pm 0.1, \pm 0.01\), provides an estimate for the derivative, \(f'(a)\).
This method of approximation involves calculating the difference quotient for values close to zero and observing that as these values get smaller, the approximations become more precise.

• Approximation simplifies complex problems and provides insight into behaviors of functions near certain points.
• In practice, technology and computational methods often rely on approximation techniques to deliver results for functions where calculating exact government is impractical.

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input. It is a basic but essential operation in mathematics, particularly in calculus.
In this exercise, evaluating the function \(f(x) = \frac{1}{x+1}\) at the point \(x=2\) and nearby points \(x = 2 \pm h\) is crucial for forming the difference quotient.
By plugging in values like \(2+0.1\) or \(2-0.1\), you calculate outputs to be used for approximating the derivative.

• Regular practice with function evaluation enhances your ability to quickly find and understand values that describe behavior of functions.
• It is also an important step in verifying the accuracy of solutions obtained through derivative approximations or other methods.