Problem 14 - Calculus for Biology and Medicine

Question

Find } c \text { so that } f^{\prime}(c)=0 . $$ $$ f(x)=x^{2}+4 x+4 $$

Step-by-Step Solution

Answer

The value of \( c \) that makes \( f'(c) = 0 \) is \( c = -2 \).

Step 1: Find the Derivative

First, we need to find the derivative of the function \( f(x) = x^2 + 4x + 4 \). The derivative of \( x^2 \) is \( 2x \), the derivative of \( 4x \) is \( 4 \), and the derivative of the constant \( 4 \) is \( 0 \). Hence, the derivative \( f'(x) = 2x + 4 \).

Step 2: Set the Derivative Equal to Zero

Next, we need to find the value of \( c \) such that \( f'(c) = 0 \). So, we set the derivative equal to zero: \( 2c + 4 = 0 \).

Step 3: Solve for c

To solve \( 2c + 4 = 0 \), subtract 4 from both sides to get \( 2c = -4 \). Then, divide both sides by 2 to solve for \( c \): \( c = -2 \).

Key Concepts

Understanding Derivatives

In calculus, the derivative of a function represents the rate at which the function's value changes. It provides us with vital information about the behavior of functions. Imagine you are tracking a car on a road. The derivative would tell you the car's speed at any point in time.

To find the derivative of a function, like the one in our problem, which is a quadratic function, we need to apply differentiation rules. For example, if you look at the function \(f(x) = x^2 + 4x + 4\), each term has its own derivative:

• The derivative of \(x^2\) is \(2x\)
• The derivative of \(4x\) is \(4\)
• The derivative of the constant \(4\) is \(0\)

Thus, combining these derivatives, we get \(f'(x) = 2x + 4\). Calculating a derivative can be one of the initial steps in many calculus problems.

Deciphering Quadratic Functions

Quadratic functions are polynomial functions of degree two. They take the general form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In our exercise, \(f(x) = x^2 + 4x + 4\) fits this form, with \(a = 1\), \(b = 4\), and \(c = 4\).

These functions have a characteristic U-shaped graph known as a parabola. The parabolas can open upwards or downwards depending on the sign of the leading coefficient "\(a\)". For our function, since \(a = 1 > 0\), the parabola opens upwards. Understanding how these functions behave in graphs helps in visualizing solutions to various problems, like finding where its derivative equals zero, which leads us to critical points.

Finding Critical Points

Critical points of a function occur where the derivative is zero or undefined. These points are crucial because they can indicate where functions have local maxima or minima (hills or valleys on the graph).

To find these points for our function \(f(x) = x^2 + 4x + 4\), we first found the derivative \(f'(x) = 2x + 4\). Then, to identify critical points, we set the derivative equal to zero: \(2x + 4 = 0\). Solving the equation helps us find the specific value of \(c\) where the rate of change transitions.

In our solution, we subtract 4 from both sides to get \(2x = -4\), and then divide by 2, resulting in \(x = -2\). Thus, \(c = -2\) is a critical point and represents where the function's slope is zero and the particular part of the function's graph lies flat momentarily.