Problem 46 - Precalculus with Calculus Previews

Question

Show that the \(x\) -intercept of the tangent line to the graph of \(f(x)=e^{x}\) at \(x=x_{\mathrm{o}}\) is one unit to the left of \(\left(x_{\mathrm{o}}, 0\right)\).

Step-by-Step Solution

Answer

The x-intercept is at \(x=x_0-1\), one unit left of \((x_0, 0)\).

Step 1: Find the slope of the tangent line

To find the slope of the tangent line to the curve \(f(x) = e^x\) at the point \(x = x_0\), we need to calculate the derivative of \(f(x)\). Since the derivative of \(e^x\) is \(e^x\) itself, the slope \(m\) of the tangent line at \(x=x_0\) is \(e^{x_0}\).

Step 2: Calculate the tangent line equation

The equation of a line in point-slope form is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. Since the point of tangency is \((x_0, e^{x_0})\), the equation becomes \(y - e^{x_0} = e^{x_0}(x - x_0)\). Simplifying, we get \(y = e^{x_0}x - e^{x_0}x_0 + e^{x_0}\).

Step 3: Find the x-intercept of the tangent line

The \(x\)-intercept occurs where \(y = 0\). Substitute \(y = 0\) into the equation from Step 2: \[0 = e^{x_0}x - e^{x_0}x_0 + e^{x_0}\]Solve for \(x\): \[e^{x_0}x = e^{x_0}x_0 - e^{x_0}\] \[x = x_0 - 1\].

Step 4: Verify the intercept condition

We have found that the \(x\)-intercept of the tangent line is at \(x = x_0 - 1\). This means the \(x\)-intercept is one unit to the left of the point \((x_0, 0)\). Thus, the condition is satisfied.

Key Concepts

Derivative

A derivative in mathematics is a fundamental concept that measures the rate at which a quantity changes. When discussing functions, the derivative represents how a function's output value changes as its input changes.
The process of finding a derivative is called differentiation. In our exercise, we need the derivative of the function \( f(x) = e^x \), an exponential function, at a specific point \( x = x_0 \).
For this function, the derivative is unique because it remains the same throughout; \( \frac{d}{dx}e^x = e^x \). This means that at any point \( x \), the slope of the tangent line to the curve is \( e^x \).

This constancy of the derivative makes exponential functions straightforward to work with when calculating the slope at any given point.

Understanding how to find derivatives is a crucial skill in calculus, helping to analyze and model real-world phenomena where change is involved.

Exponential Functions

Exponential functions are distinguished by having a constant base raised to a variable exponent, commonly written as \( a^x \) where \( a \) is a positive number. In particular, when \( a \) is Euler's number \( e \approx 2.718 \), we have the function \( f(x) = e^x \).

These functions have unique properties:

• Their rates of growth or decay increase exponentially, making them very powerful in modeling scenarios such as population growth, compound interest, and radioactive decay.
• For \( f(x) = e^x \), during differentiation, the function remains the same; thus, \( f'(x) = e^x \).

In our exercise, \( f(x) = e^x \) is key to finding the tangent line's slope at any given point, and forms the foundation for further calculations in the exercise steps.

Point-Slope Form

The point-slope form is a linear equation format used in calculus and algebra to describe the equation of a line when a point on the line and its slope are known. The formula is given by:
\[y - y_1 = m(x - x_1)\]
Where:

• \((x_1, y_1)\) is a known point on the line.
• \(m\) is the slope of the line.

Using point-slope form, we can readily find the equation of our tangent line. In the exercise, the tangent line touches the curve \( f(x) = e^x \) at the point \((x_0, e^{x_0})\).
Substituting these known values into the formula gives us:
\[y - e^{x_0} = e^{x_0}(x - x_0)\]
This formulation provides a clear and simple way to express the tangent line's equation, making it easier to find solutions like intercepts or changes between lines.

X-Intercept

An \( x \)-intercept is where a line crosses the \( x \)-axis. At this point, the value of \( y \) is zero.
This concept is crucial when analyzing tangent lines, as finding this intercept can reveal significant information about a line's orientation and positioning relative to coordinate axes.

In our given solution, once we have the equation \( y = e^{x_0}x - e^{x_0}x_0 + e^{x_0} \) from the point-slope form, we find the \( x \)-intercept by setting \( y = 0 \):\[0 = e^{x_0}x - e^{x_0}x_0 + e^{x_0} \]Solving for \( x \), we find:\[x = x_0 - 1\]This result shows that the tangent line to our curve at \( x = x_0 \) intersects the \( x \)-axis one unit to the left of \( x_0 \).
Understanding how to find the \( x \)-intercept is helpful not only in calculus but also in practical applications where such intercepts denote where certain quantities reduce to zero or change states.