Problem 16 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the four-step process to find the slope of the tangent line to the graph of the given function at any point. \(f(x)=2 x^{2}+5 x\)

Step-by-Step Solution

Answer

The short answer is: Differentiate the original function, \(f(x) = 2x^2 + 5x\), resulting in the derivative \(f'(x) = 4x + 5\). Then, the slope of the tangent line at any point \((a, f(a))\) is given by \(m = 4a + 5\). The tangent line equation is \(y - (2a^2 + 5a) = (4a + 5)(x - a)\).

Step 1: Differentiate the function with respect to x

To find the derivative, we will use the power rule of differentiation which states that if \(f(x) = ax^n\), then \(f'(x) = nax^{n-1}\). For the given function, \(f(x) = 2x^2 + 5x\), let's apply the power rule to each term: Differentiate \(2x^2\): The power rule gives us \(2 \cdot 2x^{2-1} = 4x\). Differentiate \(5x\): The power rule gives us \(5 \cdot 1x^{1-1} = 5\). Combine the differentiated terms: \(f'(x) = 4x + 5\).

Step 2: Plug the given x-value into the derivative

# Now we have the derivative, \(f'(x) = 4x + 5\). This equation gives us the slope of the tangent line at any point on the graph of the function. To find the slope of the tangent line at any desired point \((a, f(a))\), we need to plug the x-value (\(a\)) into the derivative: Slope of tangent line at \((a, f(a))\): \(m = f'(a) = 4a + 5\).

Step 3: Finding the y-coordinate of the given point

To find the y-coordinate of the given point \((a, f(a))\), we plug the x-value (\(a\)) into the original function \(f(x) = 2x^2 + 5x\): \(f(a) = 2a^2 + 5a\).

Step 4: Writing the equation of the tangent line

Now we have the point \((a, f(a))\) on the tangent line and the slope of the tangent line \(m = 4a + 5\). We can use the point-slope form of the equation of a line, which is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the line and \(m\) is the slope of the line. In this case, \((x_1, y_1) = (a, f(a))\) and \(m = 4a + 5\). Plug these values into the point-slope form equation: \(y - (2a^2 + 5a) = (4a + 5)(x - a)\). This equation represents the tangent line to the graph of the function \(f(x) = 2x^2 + 5x\) at any point \((a, f(a))\).

Key Concepts

Slope of Tangent Line

A tangent line to a curve at a given point is a straight line that just touches the curve at that point. The slope of this tangent line is an important feature as it tells us how steep the line is at the very point it touches the curve. In mathematical terms, the slope of the tangent line at a point on a function’s graph represents the rate of change of the function's value with respect to changes in the input (x-value).

To find the slope of the tangent line to a function, we use the function's derivative. The derivative, evaluated at a specific point, gives us the slope of the tangent line at that point. For instance, if you have a derivative function, say, \( f'(x) = 4x + 5 \), you can find the slope of the tangent line at any x by simply plugging that x-value into the derivative. For example, to find the slope at \((a, f(a))\), you calculate \( f'(a) \), which will give you \( 4a + 5 \).

This process is fundamental in calculus, as it helps to solve real-world problems like predicting the speed of an object at a certain time or analyzing cost functions in economics.

Power Rule of Differentiation

The power rule of differentiation is a fundamental tool for finding derivatives, especially when dealing with polynomial functions. This rule simplifies the process of differentiation, making it easier to handle functions of the form \( ax^n \).

The power rule states that if \( f(x) = ax^n \), then the derivative \( f'(x) = nax^{n-1} \). Let's break that down:

• "a" is the coefficient and remains as a constant.
• "n" is the exponent and you bring it down in front of the coefficient.
• You reduce the exponent by one to find the new power of x.

For the function \( f(x) = 2x^2 + 5x \), using the power rule:

• The derivative of \( 2x^2 \) is \( 4x \) because you take 2 times the exponent 2, which gives 4, and then subtract 1 from the exponent leading to \( x^{1} \).
• For \( 5x \), consider it as \( 5x^1 \), so the derivative is 5, since the power of x becomes zero and \( x^0 \) is 1.

This results in the derivative \( f'(x) = 4x + 5 \), allowing us to easily determine the slope of the tangent line using any x-value.

Equation of a Tangent Line

Once you know where the tangent line touches the curve and its slope at that point, you can write the equation of the tangent line. The point-slope form of a linear equation comes in handy here. This form is given by: \( y - y_1 = m(x - x_1) \)

Here is how you use it:

• \( m \) represents the slope of the tangent line.
• \( (x_1, y_1) \) is the point on the line where you want the tangent line to touch the function graph; usually, this is \( (a, f(a)) \).

To illustrate, if you are calculating the tangent line for the function \( f(x) = 2x^2 + 5x \) at the point \( (a, f(a)) \), with the tangent's slope \( 4a + 5 \), plug these values into the point-slope form:

\( y - (2a^2 + 5a) = (4a + 5)(x - a) \).

This shows how the tangent line interacts with the graph. It allows us to make approximations and understand local behavior of the function at a given point, which is powerful in both theoretical and applied mathematics.