Problem 31 - Precalculus: Mathematics for Calculus

Question

Tangent Lines (a) If \(f(x)=x^{3}-2 x+4,\) find \(f^{\prime}(a)\) (b) Find equations of the tangent lines to the graph of \(f\) at the points whose \(x\) -coordinates are \(0,1,\) and 2 (c) Graph \(f\) and the three tangent lines.

Step-by-Step Solution

Answer

The derivatives at points 0, 1, 2 are -2, 1, 10 respectively.

Step 1: Find the Derivative

To find the derivative of the function \(f(x) = x^3 - 2x + 4\), we use the power rule. The derivative is given by \[f'(x) = 3x^2 - 2.\] This expression can now be used to find the slope of the tangent line at any point \(x = a\). Thus, the expression for \(f'(a)\) is \[f'(a) = 3a^2 - 2.\]

Step 2: Calculate Slopes at Given Points

Plug the specific \(x\)-coordinates into the derivative to find the slopes of the tangents:- For \(x = 0\), \(f'(0) = 3(0)^2 - 2 = -2\).- For \(x = 1\), \(f'(1) = 3(1)^2 - 2 = 1\).- For \(x = 2\), \(f'(2) = 3(2)^2 - 2 = 10\).

Key Concepts

Derivative

The concept of a derivative is central to calculus. It represents the rate of change or the slope of a function at a given point. To find the derivative of a function, we differentiate it, which essentially means finding a new function that gives us the slope of the original function at any given point. This slope can also be thought of as the slope of the tangent line to the curve at a specific point on the graph.

In our particular exercise, we have the function \(f(x) = x^3 - 2x + 4\). When we take the derivative of this function, we get \(f'(x) = 3x^2 - 2\). This derivative function \(f'(x)\) now enables us to calculate the slope of the tangent line at any point \(x = a\). By simply substituting \(a\) into \(f'(x)\), we obtain \(f'(a) = 3a^2 - 2\), the exact slope at that particular point on the curve.

Slope of Tangent Line

The slope of a tangent line to a curve at a given point is crucial for understanding how the function behaves at that point. This slope is the same as the derivative of the function at that point. It indicates whether the function is increasing or decreasing, and by how much.

In practice, to find this slope, we evaluate the derivative at a specific point, which we did in our solution. For example:

• When \(x = 0\), the slope is \(f'(0) = -2\). This tells us the tangent line is decreasing at this point.
• When \(x = 1\), the slope is \(f'(1) = 1\). Here the tangent line is increasing slightly.
• When \(x = 2\), the slope is \(f'(2) = 10\), indicating a steep increase.

The slope of the tangent line gives us insight into how steeply or gently a function climbs or descends at particular points, guiding us when sketching or analyzing graphs.

Power Rule

The power rule is a fundamental technique in calculus used to find derivatives of functions with powers of \(x\). It makes the differentiation process straightforward, particularly for polynomial functions.

The power rule states that if you have a function \(f(x) = x^n\), its derivative is \(f'(x) = nx^{n-1}\). You multiply the power \(n\) by the function, then subtract one from the power.

Applying this to our function \(f(x) = x^3 - 2x + 4\):

• The derivative of \(x^3\) using the power rule is \(3x^2\).
• The derivative of \(-2x\) is simply \(-2\), as the power of \(x\) is 1, making it \(-2 \times 1 = -2x^{0}\) or just \(-2\).
• The derivative of a constant \(4\) is \(0\) since constants do not change.

Combining these results, the derivative is \(3x^2 - 2\). The power rule is a handy shortcut, especially for polynomials, saving time while ensuring accuracy.