Problem 73 - Essential Calculus

Question

Assume that \(f\) is differentiable for all \(x\). The signs of \(f^{\prime}\) are as follows. \(f^{\prime}(x)>0\) on \((-\infty,-4)\) \(f^{\prime}(x)<0\) on (-4,6) \(f^{\prime}(x)>0\) on \((6, \infty)\) Supply the appropriate inequality for the indicated value of \(c\). $$ g(x)=f(x)+5 \quad g^{\prime}(0) $$

Step-by-Step Solution

Answer

The derivative of function \(g\) at \(x = 0\) which is \(g'(0)\) is less than zero, i.e., \(g'(0) < 0\).

Step 1: Understand the given derivative conditions

The problem states that the derivative of \(f\) \(f'(x)\) is positive in intervals \((-\infty, -4)\) and \((6, \infty)\) and negative on \((-4, 6)\). The inequality sign changes at -4 and 6, which indicates that the function \(f\) has a local maximum at \(x = -4\) and a minimum at \(x = 6\).

Step 2: Analyze the function \(g(x) = f(x) + 5\)

The function \(g(x) = f(x) + 5\) is simply the function \(f(x)\) shifted vertically upwards by 5 units. This shift does not affect the derivative, as the derivative of a constant is zero. Therefore, \(g'(x) = f'(x)\).

Step 3: Find \(g'(0)\)

Now, you want to find the value of \(g'(0)\). We know that \(g'(x) = f'(x)\), so \(g'(0) = f'(0)\). From our information about the derivative \(f'(x)\), we know that \(f'(x)<0\) on \(-4 < x < 6\). The number 0 falls within this interval, thus \(f'(0) < 0\) and consequently \(g'(0) < 0\).

Key Concepts

Derivative

The derivative of a function gives us valuable information about its behavior, especially about how it slopes or changes. Imagine a road that goes uphill or downhill – the slope of this road at any point is what a derivative represents. Mathematically, the derivative of a function \( f(x) \) is noted as \( f'(x) \). In this exercise, the problem analyzes where the derivative is positive or negative.

A positive derivative \( f'(x) > 0 \) means that the function is increasing in that interval, like a road going uphill. Conversely, a negative derivative \( f'(x) < 0 \) indicates that the function is decreasing, or going downhill. This is how we identified that the function \( f(x) \) had certain sections where it was increasing or decreasing. Such insights hint at the presence of local maxima and minima, explored further in the coming sections.

Overall, derivatives are powerful tools to understand a function's dynamics just by investigating its slope changes. They help us anticipate and describe the behavior of functions with great precision.

Local Maximum

A local maximum of a function is a point where the function peaks – it is higher than all nearby points. You can picture it as the top of a hill. In the context of derivatives, a local maximum occurs at a point where the derivative changes from positive to negative.

In the given exercise, since \( f'(x) \) transitions from positive on \((-fty, -4)\) to negative on \((-4, 6)\), this indicates a local maximum at \( x = -4 \). It's like reaching the highest point before starting to descend downwards.

Identifying local maxima is particularly useful in optimizing real-world problems, such as identifying peak points of profit or efficiency in various processes.

Local Minimum

A local minimum is the opposite of a local maximum – it's a valley or the lowest point in a nearby range. This point will be lower than all the points surrounding it. Think of it as the bottom of a trough. In terms of derivatives, a local minimum is found where the derivative changes from negative to positive.

In our exercise, this occurs at \( x = 6 \) because the derivative shifts from negative on \((-4, 6)\) to positive on \((6, \infty)\). It's like walking down a hill and reaching the lowest point before starting to climb up again.

Recognizing local minima is vital in seeking solutions that involve minimizing cost, time, or other quantities in various fields.

Vertical Shift

The concept of vertical shift revolves around moving the graph of a function up or down without altering its shape. When a constant is added or subtracted from a function, it causes this vertical shift.

In the exercise, the function \( g(x) = f(x) + 5 \) represents a vertical shift of the function \( f(x) \) upwards by 5 units. This means every point on the graph of \( f(x) \) is simply lifted 5 units higher.

What’s important is that this vertical shift does not affect the derivative. Because the derivative of a constant (5, in this case) is zero, the slope of the function remains the same. Thus, in terms of our task, \( g'(x) = f'(x) \) everywhere.

Vertical shifts are fundamental in adjusting functions without altering their inherent behavior, often used in fine-tuning models and representations across mathematics and science.