Problem 13 - Calculus for Biology and Medicine

Question

Find } c \text { so that } f^{\prime}(c)=0 . $$ $$ f(x)=x^{2}-6 x+9 $$

Step-by-Step Solution

Answer

The value of \( c \) is 3.

Step 1: Find the Derivative

To find where the derivative equals zero, we first calculate the derivative of the function. Given the function \( f(x) = x^2 - 6x + 9 \), apply the power rule to obtain the derivative: \( f'(x) = 2x - 6 \).

Step 2: Set the Derivative Equal to Zero

To find \( c \) where \( f'(c) = 0 \), set the derivative equal to zero: \( 2x - 6 = 0 \).

Step 3: Solve for \( x \)

Solve the equation \( 2x - 6 = 0 \) for \( x \). Add 6 to both sides: \( 2x = 6 \). Then divide both sides by 2: \( x = 3 \). Therefore, \( c = 3 \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to one of its variables. Imagine you are driving a car and want to know how your speed changes over time—that change in speed is similar to what a derivative tells us about a function. In mathematics, when we talk about derivatives, we're typically looking to find the slope of a function at any given point. This slope tells us how steep the function is at that particular point. For a function like our example, the derivative is expressed mathematically.

• The derivative gives the slope of the tangent line at any point on the function.
• When the derivative equals zero, it indicates a point where the slope is flat, often referred to as a critical point.
• Critical points are essential because they help identify maximum, minimum, or saddle points of a function.

In our exercise, finding where the derivative equals zero (a critical point) is key to determining where the rate of change stops, indicating a peak or trough in the function's graph.

Exploring Quadratic Functions

A quadratic function is a polynomial function of degree two, and it has a standard form of:
\[ f(x) = ax^2 + bx + c \]This type of function graphs as a parabola, which can open upwards or downwards depending on the leading coefficient, "a".

• In an upward-opening parabola, the lowest point is the minimum, while in a downward-opening parabola, the highest point is the maximum.
• These points, called vertices, are crucial as they often occur at the critical points calculated using derivatives.
• The function given, \( f(x) = x^2 - 6x + 9 \), is a quadratic function with a simplified form showing it opens upwards (because the coefficient of \( x^2 \) is positive).

The goal in examining a quadratic function in this context is often to find where the derivative equals zero. This point represents the vertex of the parabola. In our provided function, setting the derivative to zero allowed us to find that vertex, which is the critical point \( c = 3 \). This tells us about the behavior of the function around this point.

Applying the Power Rule

The power rule is a straightforward, valuable tool in calculus used to find derivatives of functions where the variable has an exponent.
Specifically, the power rule states that for any function \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \). This means:

• Take the exponent "n" of the original term and multiply it by the term itself.
• Decrease the original exponent by one for the new exponent of the derivative.
• The power rule is easy to remember and apply to all polynomial terms.

In our specific function, \( f(x) = x^2 - 6x + 9 \), we apply the power rule to each term:

• For \( x^2 \), the derivative becomes \( 2x \) since \( (2)x^{2-1} = 2x \).
• The derivative of \( -6x \) becomes \( -6 \), as it simplifies by applying the rule (with exponent 1: \( (1)(-6)x^{1-1} = -6 \)).
• The constant term \( 9 \) disappears because the derivative of a constant is zero.

These steps give us the derivative \( f'(x) = 2x - 6 \), which is critical in solving our original problem, as it provides the information needed to find the zero point for the derivative, yielding the critical point \( c = 3 \). By mastering the power rule, one can easily tackle similar derivative problems in calculus.