Problem 42 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the slope and an equation of the tangent line to the graph of the function \(f\) at the specified point. \(f(x)=-\frac{5}{3} x^{2}+2 x+2 ;\left(-1,-\frac{5}{3}\right)\)

Step-by-Step Solution

Answer

The slope of the tangent line to the graph of the function \(f(x)=-\frac{5}{3} x^{2}+2 x+2\) at the point \((-1, -\frac{5}{3})\) is \(\frac{16}{3}\). The equation of the tangent line at this point is \(y = \frac{16}{3}(x + 1) - \frac{5}{3}\).

Step 1: Write down the function

We have the function \(f(x)=-\frac{5}{3} x^{2}+2 x+2\) and the given point \((-1, -\frac{5}{3})\).

Step 2: Calculate the derivative

To find the derivative, we will use the power rule: \(f'(x)= n \cdot x^{n-1}\). Differentiating each term gives us: \[ f'(x) = -\frac{5}{3}\cdot 2x^{2-1} + 2 \cdot x^{1-1}\\ f'(x) = -\frac{10}{3}x + 2 \]

Step 3: Find the slope at the given point

We now need to find the slope of the tangent line at the given point \((-1, -\frac{5}{3})\). To do this, we will plug the x-coordinate of the point into \(f'(x)\): \[ f'(-1) = -\frac{10}{3}(-1) + 2\\ f'(-1) = \frac{10}{3} + 2\\ f'(-1) = \frac{10}{3} + \frac{6}{3}\\ f'(-1) = \frac{16}{3} \] At the given point, the slope of the tangent line is \(\frac{16}{3}\).

Step 4: Find the equation of the tangent line

Now, we will use the point-slope form to find the equation of the tangent line: \[ y-y_{1} = m(x-x_{1}) \] Here, \(m\) is the slope, and \((x_{1}, y_{1})\) are the coordinates of the given point: \[ y - \left(-\frac{5}{3}\right) = \frac{16}{3}(x - (-1)) \] Now solve for \(y\): \[ y + \frac{5}{3} = \frac{16}{3}(x + 1)\\ y = \frac{16}{3}(x + 1) - \frac{5}{3} \] Thus, the equation of the tangent line to the graph of the function at the specified point is \(y = \frac{16}{3}(x + 1) - \frac{5}{3}\).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it, reflecting the instantaneous direction of the curve at that point. Imagine a car driving along a winding road. At any specific point on that road, the tangent line would represent the direction the car is heading. This concept is essential in calculus because it helps in approximating the behavior of curves at given points.

• The tangent line meets the curve at precisely one point.
• It shares the same slope as the curve at this point.
• This slope indicates how steep or flat the curve is.

To find an equation for the tangent line, we need both the slope at the point and the point itself. Once we have these, we can use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point.

Derivative

The derivative of a function provides a tool for determining the slope of the tangent line to the function’s graph at any point. By definition, it measures the rate at which the function's value changes as its input changes. In simpler terms, the derivative tells us how fast the function is "moving" at each point.

• The process of finding a derivative is called differentiation.
• The derivative is denoted as \(f'(x)\) or \(\frac{dy}{dx}\), among other notations.
• It can be thought of as the "instantaneous rate of change" of the function.

Calculating the derivative involves using various rules, one of the simplest being the power rule. The derivative is crucial not only for finding slopes but also for understanding the behavior of functions, especially in optimization and motion.

Slope of a Curve

The slope of a curve at a particular point indicates how steep the curve is at that point, equivalent to the slope of the tangent line there. This slope guides us in understanding the rate at which the curve is increasing or decreasing as we move from left to right.

• At a high positive slope, the curve rises steeply.
• At a zero slope, the curve flattens, possibly indicating a maximum or minimum.
• A negative slope means the curve is falling.

In our exercise, when we calculated the slope of the curve at \((-1, -\frac{5}{3})\), we plugged the x-value into the derivative, giving us the slope of \(\frac{16}{3}\). This positive result means the curve is rising sharply at that point.

Power Rule

The power rule is a straightforward technique for finding the derivative of functions with terms of the form \(ax^n\). It states that the derivative of a term \(ax^n\) is \(n \cdot ax^{n-1}\), effectively reducing the power of \(x\) by one and multiplying by the original power. This makes differentiation much more manageable.

• It simplifies the process for polynomials.
• Applicable when the exponent is a whole number, a fraction, or negative.
• Can be extended to expressions with several terms, applying to each term individually.

In the given exercise, the power rule was utilized to differentiate each part of the function \(f(x)=-\frac{5}{3} x^{2}+2 x+2\). By doing so, it transformed the function into a more accessible form, allowing the calculation of the derivative needed for determining the slope of the tangent line.