Problem 69 - College Algebra and Calculus: An Applied Approach

Question

Use the given information to find \(f^{\prime}(2)\) \(g(2)=3\) and \(g^{\prime}(2)=-2\) \(h(2)=-1 \quad\) and \(\quad h^{\prime}(2)=4\) $$ f(x)=2 g(x)+h(x) $$

Step-by-Step Solution

Answer

The value of \(f^{\prime}(2)\) is 0.

Step 1: Understand the Sum Rule of Differentiation

The sum rule in differentiation states that the derivative of the sum of two functions is the sum of the derivatives of each function. Specifically, if \(f(x) = g(x) + h(x)\), then \(f^{\prime}(x) = g^{\prime}(x) + h^{\prime}(x)\). We will apply this rule to differentiate the function \(f(x) = 2g(x) + h(x)\).

Step 2: Differentiate \(f(x)\) Using the Sum Rule

By applying the sum rule, the derivative of \(f(x) = 2g(x) + h(x)\) becomes \(f^{\prime}(x) = 2g^{\prime}(x) + h^{\prime}(x)\).

Step 3: Substitute the Given Values

Now, we use the given values for \(g^{\prime}(2)\) and \(h^{\prime}(2)\) in the equation from Step 2 to calculate \(f^{\prime}(2)\). These are \(g^{\prime}(2) = -2\) and \(h^{\prime}(2) = 4\), which makes \(f^{\prime}(2) = 2(-2) + 4 = -4 + 4 = 0\).

Key Concepts

Derivatives

Understanding derivatives is fundamental in calculus, as they represent the rate at which a function is changing at any given point. In essence, the derivative of a function at any given point reflects the slope of the tangent line to the function's graph at that point. Mathematically, if you have a function denoted by \(f(x)\), the derivative of this function is often written as \(f'(x)\) or \(\frac{df}{dx}\).

When considering our specific exercise, we want to determine the derivative of the function \(f(x)\) at the point where \(x=2\). This is represented as \(f'(2)\) and shows the slope of \(f(x)\) at that particular point. Derivatives have numerous applications, from physics to economics, anywhere a changing rate or slope needs to be quantified, derivatives are the tool of choice.

Differentiation Techniques

Differentiation techniques are the methods used to calculate the derivative of a function. One of the most fundamental techniques is the sum rule of differentiation, which is beautifully demonstrated in our exercise.

According to the sum rule, if you have two functions \(g(x)\) and \(h(x)\), and a new function \(f(x)\), which is the sum of them – that is, \(f(x) = g(x) + h(x)\) – then the derivative of \(f(x)\) is simply the sum of the derivatives of \(g(x)\) and \(h(x)\). But what if we multiply one of our functions by a constant, as we see with \(2g(x)\) in the exercise? This introduces us to another rule: the constant multiple rule. This rule states that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function, or \(\frac{d}{dx}[c\cdot g(x)] = c\cdot g'(x)\). Apply both these rules together, and differentiation becomes a more manageable process.

Calculus Applications

Calculus, particularly through the use of derivatives, plays a crucial role in various real-world applications. One of the key uses is in optimizing functions, which can help in finding maximum profit, minimizing cost, or optimizing the shape of an object for specific performance, such as aerodynamics. In physics, derivatives describe motion, allowing us to calculate velocity and acceleration from the equations of position over time.

In our textbook problem, while the context isn’t specified, the process we engage in—differentiating a function constructed from real-world data—mirrors the steps taken in countless practical situations. A deep understanding of rules like the sum rule of differentiation not only helps resolve textbook exercises but equips students with the skills to apply calculus to complex problems they might encounter in fields like engineering, computer science, biology, and economics.