Question

A function \(f\) and an \(x\) -value \(a\) are given. Approximate the equation of the tangent line to the graph of \(f\) at \(x=a\) by numerically approximating \(f^{\prime}(a),\) using \(h=0.1 .\) $$f(x)=e^{x}, x=2$$

Short answer

Tangent line at \( x=2 \) is \( y = 7.3891 + 7.771 \cdot (x - 2) \).

Step-by-step solution

Understand the given function

We are given the function \( f(x) = e^x \) which is the natural exponential function. The task is to find the equation of the tangent line at a specific point \( x = 2 \).

Recall the formula for the derivative approximation

To approximate the derivative at a point, we can use the difference quotient: \( f^{\prime}(a) \approx \frac{f(a+h) - f(a)}{h} \), where \( h \) is a small value.

Set up the derivative approximation

Set \( a = 2 \) and \( h = 0.1 \). Calculate \( f(2 + 0.1) \) and \( f(2) \). The formula for approximation becomes: \[ f^{\prime}(2) \approx \frac{f(2.1) - f(2)}{0.1} \].

Evaluate the function at given points

Calculate \( f(2) = e^2 \) and \( f(2.1) = e^{2.1} \). Using a calculator, approximate these as \( f(2) \approx 7.3891 \) and \( f(2.1) \approx 8.1662 \).

Compute the approximate derivative

Use the function values to approximate the derivative:\[f^{\prime}(2) \approx \frac{8.1662 - 7.3891}{0.1} = 7.771 \].

Write equation of the tangent line

The equation of a tangent line is given by \( y = f(a) + f^{\prime}(a) \cdot (x - a) \). Substituting \( f(2) \approx 7.3891 \) and \( f^{\prime}(2) \approx 7.771 \):\[ y = 7.3891 + 7.771 \cdot (x - 2) \].

Key concepts

Numerical Derivative

Calculating a derivative numerically is a useful way to approximate the rate at which a function changes at a particular point. This approach is especially handy when an analytical solution is difficult or impossible to obtain. In the context of our function,

• The numerical derivative involves calculating the slope of the tangent line to the curve of a function at some given point.
• We achieve this by using a small increment, or step size, denoted by "h", to measure how much the function value changes over this interval.

For example, using the formula for a numerical derivative, we can approximate the derivative of our function at the point by \[f'(a) \approx \frac{f(a + h) - f(a)}{h}\].This approach gives us a practical way to find the slope of the tangent line which represents the derivative of the function at the point of interest.

Exponential Function

The exponential function, particularly the one expressed as \( f(x) = e^x \), is fundamental in mathematics. It's known for its unique properties:

• It grows very rapidly as \( x \) increases, reflecting exponential growth.
• The base \( e \) is an irrational constant approximately equal to 2.71828, and it's often used in natural growth contexts.

In our exercise, the function is evaluated at \( x = 2 \) and \( x = 2.1 \). These specific values illustrate the nature of exponential growth over small intervals, which can be significant. The rapid growth characteristic is critical when calculating derivatives, as it influences the steepness of the tangent line.

Difference Quotient

The difference quotient is the core tool for approximating derivatives in numerical methods. It essentially provides a means to calculate the slope of the secant line between two points on a curve, which approaches the slope of the tangent line as the interval shrinks.This method is done by selecting a small \( h \) such as 0.1, as used in our example and applying it to the point of interest:\[f'(a) \approx \frac{f(a+h) - f(a)}{h}\].

• This formula measures the average rate of change of the function over the interval \([a, a+h]\).
• As \( h \) becomes smaller, the approximation becomes more precise.

While the difference quotient is an approximation, it becomes very reliable with appropriately small \( h \), making it a valuable asset in calculus.

Equation of a Tangent Line

The equation of a tangent line provides a linear approximation of a function at a specific point, capturing the essence of the function's behavior around that point. The general form of this equation is: \[y = f(a) + f'(a)\cdot(x-a)\],where \( f(a) \) is the function value at the point, and \( f'(a) \) is the derivative or slope of the function at that point.

• This line effectively "touches" the function at one point and closely follows its slope at that location.
• It's particularly useful in making predictions or understanding function behavior near the point.

This linearization technique is not only useful but a cornerstone concept in calculus, aiding in both theoretical and practical applications.