Problem 33 - Calculus

Question

True or false, when \(f(x)\) is an antiderivative of \(v(x)\) (a) \(2 f(x)\) is an antiderivative of \(2 v(x)\) (try examples) (b) \(f(2 x)\) is an antiderivative of \(v(2 x)\) (c) \(f(x)+1\) is an antiderivative of \(v(x)+1\) (d) \(f(x+1)\) is an antiderivative of \(v(x+1)\) (e) \((f(x))^{2}\) is an antiderivative of \((v(x))^{2}\).

Step-by-Step Solution

Answer

(a) True, (b) False, (c) False, (d) True, (e) False.

Step 1: Understanding Antiderivatives

An antiderivative of a function, say \(v(x)\), is a function \(f(x)\) such that \(f'(x) = v(x)\). With this in mind, we'll evaluate each statement by examining the derivatives of the given functions.

Step 2: Testing Statement (a)

For statement (a), consider \(2f(x)\). The derivative is \(\frac{d}{dx}[2f(x)] = 2f'(x)\). Since \(f'(x) = v(x)\), it follows that \(2f'(x) = 2v(x)\). This statement is true.

Step 3: Testing Statement (b)

For statement (b), consider \(f(2x)\). The derivative is obtained using the chain rule: \(\frac{d}{dx}[f(2x)] = f'(2x) \cdot 2\). Since \(f'(x) = v(x)\), we have \(f'(2x) = v(2x)\), leading to \(2v(2x)\), not \(v(2x)\). Thus, this statement is false.

Step 4: Testing Statement (c)

For statement (c), consider \(f(x) + 1\). The derivative is \(\frac{d}{dx}[f(x) + 1] = f'(x) + 0 = f'(x)\). Since \(f'(x) = v(x)\), \(f(x) + 1\) is an antiderivative of \(v(x)\), not \(v(x) + 1\). Thus, this statement is false.

Step 5: Testing Statement (d)

For statement (d), consider \(f(x+1)\). The derivative is \(\frac{d}{dx}[f(x+1)] = f'(x+1)\). Since \(f'(x) = v(x)\), we have \(f'(x+1) = v(x+1)\), meaning this statement is true.

Step 6: Testing Statement (e)

For statement (e), consider \((f(x))^2\). The derivative is \(\frac{d}{dx}[(f(x))^2] = 2f(x)f'(x)\). Since \(f'(x) = v(x)\), the result is \(2f(x)v(x)\), not \((v(x))^2\). Thus, this statement is false.

Key Concepts

Calculus

Calculus is a branch of mathematics that delves into the study of change. It is fundamental to understanding a variety of mathematical concepts including derivatives and integrals. These concepts are crucial for calculating rates of change and areas under curves.

A key goal of calculus is to find antiderivatives and derivatives of functions. An antiderivative is essentially the reverse process of differentiation, allowing us to determine a function from its derivative.

In exercises involving antiderivatives, it's common to evaluate given statements by differentiating them to see if they match up with the original function.

• Antiderivatives can have multiple forms due to the constant of integration.
• The process of finding antiderivatives helps solve differential equations, a core application of calculus.

By mastering these concepts, students can better understand the mechanics of how functions behave over intervals and at specific values.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It simplifies the differentiation process when you have a function inside another function.

For instance, if you have a function expressed as \(f(g(x))\), the chain rule states that its derivative is \(f'(g(x)) imes g'(x)\). This means you need to multiply the derivative of the outer function by the derivative of the inner function.

• This rule is essential when dealing with functions like \(f(2x)\) or \(f(x+1)\).
• The chain rule ensures accurate differentiation for these nested functions.

When students encounter a statement like statement (b) in the exercise, the chain rule clarifies why the antiderivative shifts due to the inner function's derivative influence.

Derivative

The concept of a derivative is all about understanding how a function changes at any given point. In calculus, deriving a function helps us determine the rate at which a function's value is changing.

Denoted by \(f'(x)\), the derivative provides significant insight into the properties of functions by capturing the slope of the tangent line at any point on the function's graph.

• It's crucial for evaluating and verifying antiderivatives.
• Determining whether given functions are antiderivatives involves finding their derivatives and matching them with the original functions.

In the exercise, understanding derivatives is key to deciding the truth of each statement, as it directly tests if one function is genuinely the antiderivative of another.

Function Evaluation

Function evaluation involves plugging specific values into a function to understand its behavior at certain points. This practice is critical in verifying the accuracy of derivatives and antiderivatives.

Evaluating functions helps determine the correctness of proposed antiderivatives by checking if they match the original functions upon differentiation.

Through function evaluation:

• One can verify mathematical statements by checking if the derived functions correctly map back to their respective output values.
• Statements like (d) can be assessed by seeing if functions like \(f(x+1)\) yield the correct antiderivative behavior when evaluated and differentiated.

This assessment ensures clarity and accuracy, especially when determining the true nature of a function's relationship to another.