Problem 10 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The value of \( c \) is 0.

Step 1: Differentiate the Function

The first step is to find the derivative of the given function, \( f(x) = -x^2 + 4 \). The derivative, \( f'(x) \), represents the slope of the function at any point \( x \). For \( f(x) = -x^2 + 4 \), the derivative is \( f'(x) = -2x \).

Step 2: Set the Derivative Equal to Zero

To find the value of \( c \) such that \( f'(c) = 0 \), we set the derivative function \( f'(x) = -2x \) equal to zero. This gives us the equation \(-2x = 0\).

Step 3: Solve for \( c \)

Solve the equation from the previous step: \(-2x = 0\). Divide both sides by \(-2\) to isolate \( x \). This gives us \( x = 0 \). Thus, \( c = 0 \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point. You can think of the derivative as a way to find the slope of a function's graph at a specific point. When you differentiate a function, you get a new function that tells you the slope at every point on the original curve. This is particularly useful in finding critical points, where the function might have a maximum, minimum, or a point of inflection.

In our example, we started with the function \(f(x) = -x^2 + 4\). To find the derivative, we applied basic differentiation rules. The derivative of \(-x^2\) is \(-2x\), while the derivative of a constant like 4 is zero. Therefore, the derivative of our function is \(f'(x) = -2x\). This result lets us find the slope at any point \(x\). In this context, it helps us determine where the critical points are by setting the derivative equal to zero.

Exploring Quadratic Functions

Quadratic functions are polynomial functions of degree two, typically in the form \(ax^2 + bx + c\). They create a parabolic shape on a graph when plotted. Parabolas can open upwards or downwards depending on the sign of the coefficient \(a\). If \(a\) is positive, the parabola opens upwards; if negative, it opens downwards.

In the given exercise, the function \(f(x) = -x^2 + 4\) is a quadratic function. Here, the coefficient of \(x^2\) is \(-1\), which means the graph opens downwards. This downward-opening parabola indicates a maximum point at its vertex. Understanding the nature of a quadratic function's graph helps predict the behavior of the function, such as finding its maximum or minimum value, which is crucial in many applications, like optimizing resources or forecasting trends.

Zeroes of a Function

The zeroes of a function, also known as roots or solutions, are the points where the function equals zero. These points are critical in understanding the function's behavior, as they represent the x-values at which the function crosses or touches the x-axis.

To find the zeroes, you set the function equal to zero and solve for \(x\). In relation to the derivative, we often search for x-values where the derivative is zero. These are the critical points, indicating potential maxima, minima, or inflection points. In our solution, after differentiating, we set \(f'(x) = -2x = 0\) and solved for \(x\) to find \(x = 0\). This told us that at \(x = 0\), the slope of the tangent to the curve is zero, revealing that the graph has reached either a peak or valley at this point. Hence, finding these zeroes aids significantly in sketching the graph and understanding the overall function dynamics.