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Chapter 10: Problem 794

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

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

Step by step solution

01

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: \[f'(x) = A\]
02

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.
03

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.

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