Question

If \(\mathrm{f}(\mathrm{x})=\mathrm{Ax}+\mathrm{B}\) and \(\mathrm{f}(0)=\mathrm{f}^{\prime}(\mathrm{x})=2\) then \(\mathrm{f}(1)=\) (a) 4 (b) 2 (c) 1 (d) \(-4\)

Short answer

The correct answer is \(f(1) = 4\).

Step-by-step solution

Differentiate the function

In order to find the derivative of the function \(f(x) = Ax + B\), we will differentiate with respect to x. The derivative of \(Ax\) with respect to x is A, and the derivative of a constant (B) is 0. Therefore, the derivative of \(f(x)\) is given by: \[f'(x) = A\]

Use the given conditions to solve for A and B

We know that \(f(0) = 2\) and \(f'(x) = 2\). Using these conditions, we can solve for A and B: First condition: \(f(0) = 2\) Plug in the x-value (0) into the function: \(f(0) = A(0) + B = 2\) \(B = 2\) Second condition: \(f'(x) = 2\) Since the derivative is a constant value of A, we can equate it to the given condition: \(f'(x) = A = 2\) So, A = 2 and B = 2.

Find the value of f(1)

Now that we have found the values of A and B, we can find the value of \(f(1)\) using the linear function: \(f(x) = Ax + B\) \(f(1) = 2(1) + 2 = 4\) Therefore, \(f(1) = 4\). The correct answer is (a) 4.

Key concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. In the realm of linear functions, determining the derivative is quite straightforward. When you have a linear function like \(f(x) = Ax + B\), the derivative is simply the constant coefficient \(A\) multiplying the variable \(x\).
This simplification occurs because, in linear functions, the derivative of a constant (here, \(B\)) is zero. Thus, the derivative of \(f(x) = Ax + B\) becomes \(f'(x) = A\).
Understanding derivatives is crucial because they help us analyze how the function behaves as the input value changes. In our problem, knowing that \(f'(x) = 2\) immediately tells us that the slope of the function is constant, indicating a consistent rate of change across all values of \(x\).

Differentiation

Differentiation is the process of calculating the derivative of a function. It's an essential tool in calculus, allowing us to understand and interpret the behavior of functions. In the context of linear functions, differentiation is particularly straightforward, as linear functions simply represent straight lines.
To differentiate the function \(f(x) = Ax + B\), we follow these steps:

• The derivative of the term \(Ax\) is \(A\), because \(x\) is raised to the power of 1, and by the power rule, we subtract 1 from the power.
• The derivative of the constant \(B\) is 0, since constants do not change and therefore have no rate of change.

Consequently, the differentiation of our function gives us \(f'(x) = A\). This reflects a key characteristic of linear functions: their slope, or rate of change, is constant.

Function Evaluation

Evaluating a function involves finding the value of the function for a particular input value. This involves substituting the given input into the function equation and performing the necessary calculations.
In our exercise, we are asked to evaluate the function \(f(x) = Ax + B\) at \(x = 1\). After determining the values of \(A = 2\) and \(B = 2\), we can substitute \(x = 1\) into the function:

• \(f(1) = 2(1) + 2\)
• \(f(1) = 2 + 2\)
• \(f(1) = 4\)

Function evaluation allows us to compute the output value of a function for any specific input, providing precise insight into the function's behavior at particular points.

Constants in Linear Functions

In linear functions, constants play a pivotal role. They affect the position and orientation of the line on the Cartesian plane. Our function \(f(x) = Ax + B\) includes two constants, \(A\) and \(B\), which we solved for.

• \(A\), also called the coefficient of \(x\), determines the slope of the line. For example, \(A = 2\) implies the line rises 2 units vertically for every 1 unit it moves horizontally.
• \(B\) is the y-intercept of the line. It represents the point where the line crosses the y-axis, or the value of \(f(x)\) when \(x = 0\).

Understanding constants in linear functions helps us quickly visualize the graph of the equation and predict how changes to the constants might affect the function.