Question

Linear Approximation Consider the function \(f(x)=x^{3 / 2}\) with the solution point \((4,8) .\) (a) Use a graphing utility to graph \(f .\) Use the zoom feature to obtain successive magnifications of the graph in the neighborhood of the point \((4,8) .\) After zooming in a few times, the graph should appear nearly linear. Use the trace feature to determine the coordinates of a point near \((4,8) .\) Find an equation of the secant line \(S(x)\) through the two points.

Short answer

The secant line \(S(x)\) through the points has the equation \(y = 3x - 4\).

Step-by-step solution

Second Point Approximation

An approximate second point can be generated by using a small increment 'h' near the x-coordinate of the given point. Since the existing point is at x=4, let's take the second point to be at x=(4+h), for a tiny 'h'. Hence, the second point is \((4+h, f(4+h))\) where \(f(x) = x^{3/2}\). Subsequently, the second point becomes \((4+h, (4+h)^{3/2})\).

Slope Calculation

The slope of the secant line passing through the two points can be found by the formula \((y_2-y_1)/(x_2-x_1)\). Here, \((x_1,y_1)\) is the first point \((4,8)\) and \((x_2,y_2)\) is the second point \((4+h, (4+h)^{3/2})\). Substituting these into the formula gives us a slope \(m=(((4+h)^{3/2})-8)/(h)\). As 'h' is a very small number, after simplifying, we get \(m \approx 3\).

Secant Line Equation

The equation of a line in point-slope form is \(y - y_1 = m(x - x_1)\). Substitute \(x_1 = 4\), \(y_1 = 8\) and \(m = 3\) into this formula to get the equation of the line: \(y - 8 = 3(x - 4)\). After simplifying this equation, we obtain the final result: \(y = 3x - 4\).

Key concepts

Secant Line Equation

When students start delving into the world of calculus, the notion of a secant line emerges as a fundamental concept. A secant line is essentially a straight line that slices through a curve at two distinct points. It's an important precursor to the concept of a tangent line, which is a key player in calculus. To construct the equation of a secant line, you need two points on the function that the secant intersects.

In the exercise we're considering, those two points are \( (4,8) \) and \( (4+h, (4+h)^{3/2}) \), where 'h' represents a small increment from the x-coordinate of the first point. By knowing these two points, you're able to apply the point-slope form, which is an immensely handy tool in algebra and calculus alike. It goes like this: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the line, and \( (x_1, y_1) \) is a known point on the line. This formula not only forms the backbone for finding secant lines but also prepares students for linear approximations that are a pivotal part of calculus.

Slope Calculation

Slope is the measure of the steepness of a line, which in a geometric sense, tells us how much the line inclines or declines. In the context of our exercise, calculating the slope of the secant line is crucial. The slope formula \( (y_2-y_1)/(x_2-x_1) \) is deceptively simple, yet it encapsulates the rate of change between two points on a curve. This calculation sets the stage for understanding derivatives, which are the core of differential calculus.

To put it into practice, consider the slope \( m \) of the line connecting our known point \( (4,8) \) and our estimated second point \( (4+h, (4+h)^{3/2}) \). By substituting these into our slope formula, we reach a value that approximates the rate of change at the first point when 'h' is infinitesimally small. This calculated slope \( m \) will then be plugged into the point-slope equation, bringing us one step closer to the secant line equation that estimates the behavior of the function at the point of interest.

Increment 'h'

In calculus, the increment \( h \) might seem minor, but it's a powerful player when it comes to understanding the behavior of functions. It signifies a very small change in the x-value, nudging us from a known point to another nearby point on the curve. This incremental approach is pivotal in forming the bridge from finite differences to infinitesimal changes, the latter being the heart of calculus.

Choosing \( h \) can be a bit of an art; too large and our secant line won't accurately represent the instantaneous rate of change; too small and we risk computational errors. Nevertheless, in the grand scheme of the calculus picture, 'h' acts as our microscope, allowing us to zoom in on a function's graph and discern its characteristics with greater precision. By understanding how to manipulate and use this small increment, students gain insights into the broader concepts of limits and derivatives, which unlock the understanding of motion, growth, and a myriad other natural phenomena.