Problem 41 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 41-44, find the slope and an equation of the tangent line to the graph of the function \(f\) at the specified point. \(f(x)=2 x^{2}-3 x+4 ;(2,6)\)

Step-by-Step Solution

Answer

The slope of the tangent line to the graph of the function \(f(x) = 2x^2 - 3x + 4\) at the point \((2, 6)\) is \(5\), and the equation of the tangent line is \(y = 5x - 4\).

Step 1: 1. Find the derivative of the function.

To find the derivative of the function \(f(x) = 2x^2 -3x + 4\), we apply the power rule. The power rule states that for any function in the form of \(f(x) = ax^n\), the derivative is given by \(f'(x) = n \cdot ax^{n-1}\). Differentiating each term in the function: \(f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(4)\) Applying the power rule to each term: \(f'(x) = 4x-3\)

Step 2: 2. Evaluate the derivative at x=2.

Now, we need to find the value of the derivative at \(x = 2\). This will give us the slope of the tangent line at the specified point \((2, 6)\). \(f'(2) = 4(2) - 3 = 8 - 3 = 5\) So, the slope of the tangent line at the point \((2, 6)\) is \(5\).

Step 3: 3. Find the equation of the tangent line.

Now that we have the slope of the tangent line and the point \((2, 6)\), we can use the slope-point form of the equation for the tangent line. The slope-point form is given by: \(y-y_1 = m(x-x_1)\) where \((x_1, y_1) = (2, 6)\) and \(m = 5\). Plugging in the values, we get: \(y-6 = 5(x-2)\) Expanding the equation, we get the equation of the tangent line: \(y = 5x - 10 + 6\) \(y = 5x - 4\) Hence, the slope of the tangent line is \(5\) and the equation of the tangent line is \(y = 5x - 4\).

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus. In simple terms, it represents the rate at which a function is changing at any given point. For a function like \( f(x) = 2x^2 - 3x + 4 \), the derivative tells us how \( f(x) \) changes as \( x \) changes. To find the derivative, you apply differentiation rules, most commonly the power rule, which we'll discuss in detail.

When you find the derivative of a function, you are essentially finding a new function, \( f'(x) \), which represents the slope of the tangent line to the curve of the original function at any point \( x \). This means if you want to know how steep the function is climbing or falling at some point, the derivative is the tool you need.

Tangent Line

A tangent line to a function at a given point touches the graph at just that point without crossing it. It's like a snapshot of the slope of the curve at that particular point. For the function \( f(x) = 2x^2 - 3x + 4 \) at the point \( (2, 6) \), a tangent line provides a linear approximation of the curve near that point.

Deriving the equation of a tangent line involves finding two things: the slope of the tangent (which is the same as the value of the derivative at that point), and a specific point on the line. Using the slope-point formula, \( y - y_1 = m(x - x_1) \), you can easily calculate this. Here, \( m \) is the slope, and \( (x_1, y_1) \) is the specific point on the curve.

Power Rule

The power rule is a shorthand method to differentiate functions and is based on the principle that a function \( f(x) = ax^n \) can be easily differentiated by multiplying \( n \) by \( a \) and reducing the power by 1. In other words, the derivative \( f'(x) = n \cdot ax^{n-1} \).

Applying this to each term individually makes complex calculus operations much simpler. For the function in our exercise, \( f(x) = 2x^2 - 3x + 4 \), apply the power rule to get:

• \( 2x^2 \) becomes \( 4x \)
• \( -3x \) becomes \( -3 \)
• \( 4 \) becomes \( 0 \) as constants differentiate to zero.

This results in the derivative \( f'(x) = 4x - 3 \). The simplicity and efficiency of the power rule is one reason it is so widely used.

Slope Formula

The slope of a line measures how steep a line is. In calculus, this is particularly crucial when dealing with tangent lines, as the slope of the tangent line at a given point equals the derivative of the function at that point.

The slope formula used is derived from the derivative, and it follows that for a given point \( (x_1, y_1) \) and slope \( m \), the equation \( y - y_1 = m(x - x_1) \) describes the tangent line fully.

In our exercise, we evaluated the derivative \( f'(x) = 4x - 3 \) at \( x = 2 \) to find the slope \( m = f'(2) = 5 \). This gives us enough information to use the slope-point form of the line equation to find the tangent line at the specified point. The result is an equation \( y = 5x - 4 \), describing a line that just grazes the curve of \( f(x) \) at \( (2, 6) \). This precisely illustrates how the slope formula is employed to connect derivatives with tangible line equations.