Problem 25 - Calculus for Biology and Medicine

Question

In Problems , find c such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) $$ f(x)=(x+1)^{2} $$

Step-by-Step Solution

Answer

\( c = -1 \), and \( f(x) \) has a local minimum at this point.

Step 1: Calculation of the First Derivative

In order to find the value of \( c \), we first need to calculate the derivative of the function \( f(x) = (x+1)^2 \). Using the power rule, the derivative is \( f'(x) = 2(x+1) = 2x + 2 \).

Step 2: Setting the Derivative Equal to Zero

To find the critical points where \( f'(c) = 0 \), we set the derivative \( 2x + 2 \) equal to zero and solve for \( x \). \( 2x + 2 = 0 \Rightarrow 2x = -2 \Rightarrow x = -1 \). Thus, \( c = -1 \).

Step 3: Second Derivative Test

To determine whether \( f(x) \) has a local extremum at \( x = -1 \), we use the second derivative test. Calculate the second derivative: \( f''(x) = 2 \). Since \( f''(x) = 2 > 0 \), the function is concave up at \( x = -1 \), indicating a local minimum.

Key Concepts

Understanding the Derivative

The derivative is a fundamental concept in calculus. It represents how a function changes as its input changes. Imagine it as the slope of the function at a particular point. When you take the derivative of a function, you are finding a new function that tells you the slope of the original function at any point. The basic rule for finding the derivative of a function like \(f(x) = (x+1)^2\) is called the power rule. In this rule, you bring down the exponent in front and subtract one from the exponent. So, for \(f(x) = (x+1)^2\), the derivative is \(f'(x) = 2(x+1)\), which simplifies to \(2x + 2\). This new expression \(f'(x)\) tells us the slope of the function \(f(x)\) at any point \(x\). In the context of finding a critical point where a local extremum might exist, we often set this derivative equal to zero. Why? Because wherever the slope is zero, the function might be at a peak, a valley, or perhaps neither. These are points worth investigating.

Identifying Local Extrema

Local extrema are points on a graph where the function reaches a peak or a valley compared to nearby points. These are the local maximum or minimum values that a function can take. To find these points, you'd first look at where the derivative equals zero (as mentioned in the previous section), since a derivative of zero means a flat slope at that specific point. In our exercise, we set \(2x + 2 = 0\) and found \(x = -1\). This is a potential point where the function could have a local extremum.Once you identify these points, it's essential to determine what kind of extremum, if any, exists there. Not every point where the derivative is zero will be a maximum or a minimum spot for the function. This is where further testing is needed to be sure.

Applying the Second Derivative Test

One effective method to confirm whether a point is a local extremum is the second derivative test. The second derivative provides insight into the concavity of the function, revealing whether a curve opens upwards or downwards. When you compute the second derivative, as with our function \(f(x) = (x+1)^2\), you find \(f''(x) = 2\). Notice that this second derivative is constant and positive everywhere, particularly at our critical point \(x = -1\). With a positive second derivative, the curvature of the function is concave upward, indicating that the point is a local minimum. So for \(x = -1\), our function has a local minimum. The second derivative tells us that the slope is increasing at this point. Conversely, if the second derivative were negative, the function would be concave down, pointing to a local maximum. Thus, the sign of the second derivative directly informs us about the nature of the extremum.