Problem 31 - Calculus Volume 1

Question

Given the function \(y=f(x)\), a. find the slope of the secant line \(P Q\) for each point \(Q(x, f(x))\) with \(x\) value given in the table. b. Use the answers from a. to estimate the value of the slope of the tangent line at \(P\). c. Use the answer from b. to find the equation of the tangent line to \(f\) at point \(P\). \(f(x)=x^{2}+3 x+4, P(1,8)\) (Round to 6 decimal places.) $$ \begin{array}{|l|l|l|l|} \hline \boldsymbol{x} & \begin{array}{l} \text { Slope } \\ \boldsymbol{m}_{P Q} \end{array} & \boldsymbol{x} & \begin{array}{l} \text { Slope } \\ \boldsymbol{m}_{P Q} \end{array} \\ \hline 1.1 & \text { (i) } & 0.9 & \text { (vii) } \\ \hline 1.01 & \text { (ii) } & 0.99 & \text { (viii) } \\ \hline 1.001 & \text { (iii) } & 0.999 & \text { (ix) } \\ \hline 1.0001 & \text { (iv) } & 0.9999 & \text { (x) } \\ \hline 1.00001 & \text { (v) } & 0.99999 & \text { (xi) } \\ \hline 1.000001 & \text { (vi) } & 0.999999 & \text { (xii) } \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Tangent slope: 5; Equation: y = 5x + 3.

Step 1: Calculating Points on the Function

First, determine the function values for the points given in the table. For any given \(x\), calculate \(f(x) = x^2 + 3x + 4\). Start with \(x = 1.1\):\[ f(1.1) = (1.1)^2 + 3(1.1) + 4 = 1.21 + 3.3 + 4 = 8.51 \]and similarly calculate for all other \(x\) values.

Step 2: Slope of the Secant Line (PQ) Calculation

For each \(x\) given, calculate the slope of the secant line \(m_{PQ}\) using the formula:\[ m_{PQ} = \frac{f(x) - f(1)}{x - 1} \]For \(x = 1.1\):\[ m_{PQ} = \frac{f(1.1) - 8}{1.1 - 1} = \frac{8.51 - 8}{0.1} = 5.1 \]Repeat for all given \(x\) values in the table.

Step 3: Estimating Slope of the Tangent Line

Use the calculated slopes from the points that approach \(x = 1\) to estimate the slope of the tangent line at point \(P(1, 8)\). As \(x\) approaches 1, observe that the slopes stabilize to a specific value. The slopes calculated for the points around 1.0 suggest the slope of the tangent line is approximately \(5\).

Step 4: Equation of the Tangent Line

With the determined slope \(m\) from the tangent estimation as \(5\), and using point \(P(1, 8)\), apply the point-slope formula for a line:\[ y - y_1 = m(x - x_1) \]Substitute \(m = 5\), \(x_1 = 1\), and \(y_1 = 8\):\[ y - 8 = 5(x - 1) \]Thus, the equation of the tangent line is:\[ y = 5x + 3 \]

Key Concepts

Secant Line

Secant lines are fundamental in understanding slopes between two points on a curve. In essence, a secant line intersects a curve at two distinct points. These lines are a precursor to understanding tangent lines, as they provide a simple way to calculate an average rate of change over an interval.
In mathematical terms, the slope of a secant line connecting points \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on a curve is given by:

• \(m_{PQ} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\)

In the original exercise, this concept was demonstrated by using several values of \(x\) to compute the slopes of secant lines connecting each \(x\) to the point \(P(1, 8)\). These computations are crucial as they guide us towards identifying the slope of the tangent line.

Differentiation

Differentiation is a core concept in calculus used to find the rate at which a function is changing at any given point. Differentiating a function involves finding its derivative. This provides us with a precise algebraic expression for the slope of the tangent line at any specific point on the curve.
For instance, if we have \(f(x) = x^2 + 3x + 4\), the process of differentiation allows us to find \(f'(x)\). This derivative is calculated using:

• \(f'(x) = 2x + 3\)

Differentiation yields an instantaneous rate of change, which is what the slope of a tangent line represents—unlike a secant line that only gives an average. In context, the slope at the specific point \(P(1, 8)\) was derived by considering very close values of \(x\) to 1, observed to be approaching the derivative value.

Point-Slope Formula

The point-slope formula is a practical method used to determine the equation of a line given a point on the line and the slope. This is particularly helpful once the slope of the tangent line is known, typically from differentiation.
The point-slope formula is:

• \(y - y_1 = m(x - x_1)\)

Here, \(m\) is the slope, and \( (x_1, y_1)\) is a point on the line. In our example, the slope was estimated to be \(5\), using steps in secant lines and differentiation, while the point \(P(1, 8)\) provides the \(x_1\) and \(y_1\). By plugging these into the formula, the equation of the tangent line was determined as \(y = 5x + 3\). This equation allows us to understand how the function behaves locally around the point of tangency.