Problem 138 - Calculus Volume 1

Question

Find the equation of the tangent line to the graph of $$f(x)=x^{2}+\frac{4}{x}-10\( at \)x=8$$

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 15.9375x - 74\).

Step 1: Find the value of the function at x=8

Substitute \(x=8\) into the function \(f(x) = x^2 + \frac{4}{x} - 10\). This gives \(f(8) = 8^2 + \frac{4}{8} - 10\). Simplifying, we calculate \(f(8) = 64 + 0.5 - 10 = 54.5\).

Step 2: Calculate the derivative of the function

The derivative \(f'(x)\) of the function \(f(x) = x^2 + \frac{4}{x} - 10\) is found by differentiating each term separately: the derivative of \(x^2\) is \(2x\), the derivative of \(\frac{4}{x}\) is \(-\frac{4}{x^2}\), and the derivative of a constant \(-10\) is 0. Therefore, \(f'(x) = 2x - \frac{4}{x^2}\).

Step 3: Evaluate the derivative at x=8

Now substitute \(x=8\) into \(f'(x) = 2x - \frac{4}{x^2}\). This gives \(f'(8) = 2(8) - \frac{4}{8^2} = 16 - \frac{4}{64} = 16 - 0.0625 = 15.9375\).

Step 4: Find the equation of the tangent line

The equation of a line in point-slope form is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. From steps 1 and 3, we have \((8, 54.5)\) and the slope \(m = 15.9375\). Substituting these into the point-slope equation gives: \(y - 54.5 = 15.9375(x - 8)\). Solve for \(y\) to get the equation in slope-intercept form: \(y = 15.9375x - 74\).

Key Concepts

Tangent Line

A tangent line is an important concept in calculus. It touches a curve at just a single point, matching the curve's slope at that exact point. This means that while the tangent line and the curve share a common point, they do not intersect again, though visually, they may seem to closely follow each other.
If you imagine a curve as a road on a hill, the tangent line is like a railing pointing straight out from the road at a certain spot. It's the straight path you'd take if you continued in the same direction the road was pointing right there.
When we talk about finding a tangent line, the main target is to determine its exact position and slope at the specific point where it meets the curve. This involves understanding what's happening at that single point in terms of both position and direction.

Derivative Calculation

Calculating a derivative is like finding a formula for the slope of a curve. In the world of calculus, derivatives tell us how steep a curve is at any point. It's like speed for a car, indicating how quickly or slowly the car's position changes over time.
For the function given in the exercise, each term is differentiated separately:

• The derivative of the square term, \(x^2\), is \(2x\), indicating a doubling effect on the variable \(x\).
• For \(\frac{4}{x}\), the derivative is \(-\frac{4}{x^2}\), showing a reciprocal relationship involving squares.
• And for constants, like \(-10\), the derivative is zero, because constants do not change, so they don't affect the slope.

Once all pieces are differentiated, these are pieced together to form the entire derivative of the function. This unified equation then allows us to calculate the slope of the tangent at any point on the curve.

Point-Slope Form Equation

The point-slope form of an equation is a straightforward method to write the equation of a line when you know a point on the line and its slope. The formula is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is any known point on the line and \(m\) is its slope.
In the case of the tangent line, we use the calculated derivative value as the slope \(m\), since the derivative at a specific point gives the slope of the tangent line there. For example, at \(x = 8\), the derivative \(f'(8)\) tells us the slope at this point. The point \((8, 54.5)\) is used to indicate where exactly on the x-y plane the line should be positioned.

• Substitute \(x_1 = 8\) and \(y_1 = 54.5\) into the formula, along with the calculated slope, to get the final equation.

This gives us not only the direction but also the precise path of the tangent line on the graph.