Problem 65 - Calculus for Biology and Medicine

Question

Find the coordinates of all of the points of the graph of \(y=f(x)\) that have horizontal tangents. $$ f(x)=3 x-x^{2} $$

Step-by-Step Solution

Answer

The point with a horizontal tangent is \( \left( \frac{3}{2}, \frac{9}{4} \right) \).

Step 1: Find the Derivative of the Function

To find the points where the tangent is horizontal, we first need to find the derivative of the function since the slope of the tangent line is given by the derivative. The function given is \( f(x) = 3x - x^2 \). The derivative is calculated as follows:\[ f'(x) = \frac{d}{dx}(3x - x^2) = 3 - 2x \].

Step 2: Set the Derivative Equal to Zero

Horizontal tangents occur where the slope is zero, so we set the derivative equal to zero and solve for \( x \):\[ 3 - 2x = 0 \].

Step 3: Solve for x

To solve \( 3 - 2x = 0 \), start by adding \( 2x \) to both sides:\[ 3 = 2x \].Now, divide by 2:\[ x = \frac{3}{2} \].This is the \( x \)-coordinate where the tangent is horizontal.

Step 4: Find the Corresponding y-coordinate

Substitute \( x = \frac{3}{2} \) back into the original function to find the corresponding \( y \)-coordinate:\[ y = 3\left(\frac{3}{2}\right) - \left(\frac{3}{2}\right)^2 \].Calculate each term:\[ y = \frac{9}{2} - \frac{9}{4} \].Subtract to find \( y \):\[ y = \frac{18}{4} - \frac{9}{4} = \frac{9}{4} \].

Step 5: State the Coordinates of the Point

The coordinates of the point where the tangent is horizontal are \( \left( \frac{3}{2}, \frac{9}{4} \right) \).

Key Concepts

Derivative

In calculus, the derivative is a fundamental concept that represents the rate at which a function is changing at any given point. Essentially, it tells us the slope of the tangent line to the function's graph at each point.
For any function like our example, \( f(x) = 3x - x^2 \), finding the derivative helps us understand how the function behaves locally.

• The derivative of a term like \( 3x \) is simply the coefficient 3 because the derivative of \( x \) is 1.
• The derivative of \( x^2 \) is \( 2x \), following the power rule which states that the derivative of \( x^n \) is \( nx^{n-1} \).

This is why our derivative becomes \( f'(x) = 3 - 2x \).
Understanding derivatives is crucial for problem-solving in calculus, as they provide insights on the function's increasing or decreasing behaviors.

Horizontal tangents

Horizontal tangents are special points on a function's graph where the slope is zero. These occur at the crest or trough of a graph, often indicating a maximum or minimum point. To identify these horizontal tangents, we set the derivative equal to zero.
For the given function, we found the derivative to be \( f'(x) = 3 - 2x \).

• By setting \( 3 - 2x = 0 \), we solve for \( x \) to find points where our slope equals zero.
• Solving this equation gives \( x = \frac{3}{2} \), meaning the horizontal tangent occurs when \( x \) is \( \frac{3}{2} \).

These points are vital for analyzing and understanding the overall shape of the graph and potential turning points.

Function graph

The graph of a function provides a visual representation of all the possible values of \( f(x) \) for different \( x \) values. Plotting the graph often reveals critical features, such as where the function increases, decreases, or levels out.
For our example of \( f(x) = 3x - x^2 \), its graph is a parabola opening downwards, due to the \( -x^2 \) term.

• At \( x = \frac{3}{2} \), we identified a horizontal tangent, meaning the graph reaches a peak and the function momentarily stops increasing and starts decreasing.
• Substituting \( x = \frac{3}{2} \) into \( f(x) \) provided us the y-coordinate \( \frac{9}{4} \), helping us plot the point \( \left( \frac{3}{2}, \frac{9}{4} \right) \).

This analysis gives us an insight into the shape and behavior of the function. Knowing where horizontal tangents occur allows us to pinpoint the highest or lowest points.

Critical points

Critical points of a function are those \( x \)-values where the derivative is either zero or undefined, marking potential peaks, troughs, or points of inflection. These points are crucial for understanding and sketching the function's behavior.
In our case, since \( f'(x) = 3 - 2x \) is always defined, the critical points occur where \( f'(x) = 0 \).

• The equation \( 3 - 2x = 0 \) resolved to \( x = \frac{3}{2} \), marking this as a critical point.
• By substituting back into \( f(x) \), we calculated \( y = \frac{9}{4} \), giving us the coordinates of the critical point at \( \left( \frac{3}{2}, \frac{9}{4} \right) \).

Recognizing critical points is essential in calculus, as they inform about local maxima, minima, and the concavity of the graph, aiding deeper analysis and representation of functional behavior.