Question

In Exercises, find the slope of the graph at the indicated point. Then write an equation of the tangent line to the graph of the function at the given point. $$ f(x)=x \log _{2} x, \quad(1,0) $$

Short answer

The equation of the tangent line is \(y = \frac{1}{\ln(2)}x - \frac{1}{\ln(2)}\).

Step-by-step solution

Find the Derivative

The function \(f(x) = x \log_2(x)\) is a product of two functions, \(u(x) = x\) and \(v(x) = \log_2(x)\). So, use the product rule to find its derivative. The product rule states that \((uv)' = u'v + uv'\). Differentiate each part: \(u'(x) = 1\) and \(v'(x) = \frac{1}{x \ln(2)}\). Substitute these into the product rule to obtain \(f'(x) = 1*\log_2(x) + x*\frac{1}{x \ln(2)} = \log_2(x) + \frac{1}{\ln(2)}\).

Find the Slope at the Given Point

Now, calculate the slope at the point \(x=1\) by plugging this value into the derivative: \(f'(1) = \log_2(1) + \frac{1}{\ln(2)} = 0 + \frac{1}{\ln(2)} = \frac{1}{\ln(2)}\). This is the slope of the tangent line at that point.

Write the Equation of the Tangent Line

Use the point-slope form of a line, \(y-y_1 = m(x-x_1)\), where \((x_1, y_1)\) is the point of tangency (1,0) and \(m\) is the slope. Substituting these values, obtain the equation \(y-0 = \frac{1}{\ln(2)}(x-1)\), simplifying to \(y = \frac{1}{\ln(2)}x - \frac{1}{\ln(2)}\).

Key concepts

Tangent Line

When you have a curve on a graph, the tangent line at any point on that curve is like a straight line that just touches the curve at that very spot. It mirrors the direction the curve is heading right at that point and is "tangent" to the curve, meaning it only touches it at one spot.
This tangent line is incredibly useful because it provides insight into the nature of the curve at that particular point.
To find the equation of the tangent line, you need two things:

• The slope of the tangent line at the point of interest.
• The coordinates of the point of tangency.

Once you have these elements, you can apply the point-slope form of a linear equation. This formula is expressed as:

• \(y - y_1 = m(x - x_1)\)

where \(m\) is the slope, and \((x_1, y_1)\) is the point where the line is tangent to the curve. In our example, the tangent line at the point \((1,0)\) was found by using the slope \(\frac{1}{\ln(2)}\). The resulting equation beautifully draws a line that just grazes the curve at \(x=1\).

Product Rule

The product rule is a fundamental technique in calculus used to find the derivative of a product of two functions. This is particularly important since many functions you'll encounter are products of simpler functions. The rule can be stated as:

• If you have two functions \(u(x)\) and \(v(x)\), their product \(f(x) = u(x) \, v(x)\) has a derivative given by the formula: \( (uv)' = u'v + uv' \).

In simpler words, to find the derivative of \(f(x)\), differentiate \(u(x)\) as if \(v(x)\) wasn't there, then do the same for \(v(x)\), and finally, sum both results. In the original exercise, this method was applied to \(u(x) = x\) and \(v(x) = \log_2(x)\).
The derivatives were calculated as \(u'(x) = 1\) and \(v'(x) = \frac{1}{x \ln(2)}\), then plugged into the product rule formula to get \(f'(x) = \log_2(x) + \frac{1}{\ln(2)}\). By mastering the product rule, dissecting complex functions becomes a straightforward process.

Slope of a Function

The slope of a function gives us the steepness or incline of a curve at a point. It tells us how much the function is rising or falling as you move along the curve. Calculating the slope involves finding the derivative of the function, which represents the rate of change.
When you plug a specific value into the derivative, you find the slope at that particular point on the curve. Think of the slope as a measure of the curve's intensity at that moment.
In this exercise, finding the slope at \((1,0)\) meant substituting \(x=1\) into the derivative, yielding the value \(\frac{1}{\ln(2)}\). This value described how the function was changing at that exact point.
Using the slope at a point and the point itself, we can form the tangent line equation. Thus, understanding slopes helps connect the behavior of a function with linear approximations, bridging the world of curves and straight lines seamlessly.