Problem 46 - Precalculus with Limits

Question

In Exercises 43-50, (a) find the slope of the graph of \(f\) at the given point, (b) use the result of part (a) to find an equation of the tangent line to the graph at the point, and (c) graph the function and the tangent line. \(f(x)= x^3 - x, \quad (2, 6)\)

Step-by-Step Solution

Answer

The derivative of the given function is \(f'(x) = 3x^2 - 1\). The slope of the graph of \(f\) at the point (2, 6) is 11. The equation of the tangent line to the graph at the point (2, 6) is \(y = 11x - 16\).

Step 1: Calculation of Derivative

Differentiate the function \(f(x)= x^3 - x\) with respect to \(x\) to find the derivative of \(f\). By applying the power rule for differentiation, the derivative \(f'(x)\) is \(3x^2 - 1\).

Step 2: Find the slope at given point

Substitute the given \(x\) value from the point into the derivative to find the slope of the tangent at that point. In this case, the point is (2, 6), so \(x = 2\). Substitute \(x\) into \(f'(x)\) to get \(f'(2) = 3*2^2 - 1 = 11 \). So, the slope of the tangent line at the point (2, 6) is 11.

Step 3: Equation of the Tangent Line

Now we have the slope and the point, and can use the point-slope form of equation for a line, \( y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point on it. Substituting our slope \(m = 11\) and point \((2, 6)\) into the equation we get: \(y - 6 = 11(x - 2)\), which simplifies to \(y = 11x - 16\). So, the equation of the tangent line at the point (2, 6) is \(y = 11x - 16\).

Step 4: Graph of the Function and the Tangent Line

The graph of the function \(f(x) = x^3 - x\) and the graph of its tangent line \(y = 11x -16\) at the point (2, 6) can be sketched using any graph plotting tool. Simply input the equations and plot them onto the same graph for visualization. The function graph is a curved line while the tangent line is a straight line that touches the graph at only one point, i.e., the point (2, 6).

Key Concepts

Derivative

The derivative is a core concept in calculus, representing the rate at which a function is changing at any given point.

• To find the derivative of a function, you often use rules such as the power rule. For the function given in the exercise, which is \(f(x) = x^3 - x\), the power rule helps us derive each term separately.
• Applying the power rule, the derivative of \(x^3\) is \(3x^2\), and for \(-x\), it is \(-1\). Therefore, the derivative \(f'(x)\) is \(3x^2 - 1\).
• The derivative function \(f'(x)\) tells us the slope of the original function \(f(x)\) at any point \(x\).

Understanding the derivative is crucial to grasping how functions change and how to compute the slope of a tangent line to a curve.

Slope of a Curve

The slope of a curve at any point is determined by its derivative at that specific point. This slope represents the steepness or incline of the curve at a particular location.

• To find the slope at a given point, you substitute the \(x\) value of the point into the derivative function \(f'(x)\). For example, with the point \( (2, 6) \) from our exercise, you substitute \(x = 2\) into \(f'(x) = 3x^2 - 1\).
• When \(x = 2\), the computation gives us \(f'(2) = 3(2)^2 - 1 = 11\). This means the slope of the curve \(f(x) = x^3 - x\) at the point (2,6) is 11.

This slope is critical when drawing tangent lines since it indicates how the curve slopes at that particular point, functioning as the tangent line's slope.

Point-Slope Form

Point-slope form is a method for writing the equation of a line that's very handy when you have a point and the slope of the line.This form is described by the equation:

• \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the known point on the line.

Given the exercise, once you've determined that the slope \(m\) is 11 and the point is \((2, 6)\), it's straightforward to plug these into the point-slope formula:

• \(y - 6 = 11(x - 2)\).
• Then, you simplify it to obtain the line's equation in slope-intercept form: \(y = 11x - 16\).

The point-slope form is very versatile, allowing you to quickly write equations of lines without needing to perform extensive calculations when you know a single point and the line's slope.