Question

Let $f:R \rightarrow R$ be given by $f(x + y) = f(x) – f(y) + 2xy + 1$ for all $x, y \in R$. If $f(x)$ is everywhere differentiable and $f'(0)=1$, then $f'(x)$ is equal to

• A. 2x + 1
• B. 2x-1
• C. x + 1
• D. x-1

Answer 1

Answer: (B)

2x-1

Answer 2

The given functional equation is $f(x + y) = f(x) – f(y) + 2xy + 1$ for all $x, y \in R$. The function $f(x)$ is everywhere differentiable, and $f'(0)=1$. We need to find $f'(x)$.

Method 1: Using the definition of the derivative.

The definition of the derivative of $f(x)$ is $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. From the given functional equation, let $y=h$. We have: $f(x+h) = f(x) - f(h) + 2xh + 1$. Rearranging this equation to get $f(x+h) - f(x)$: $f(x+h) - f(x) = -f(h) + 2xh + 1$.

Substitute this into the definition of $f'(x)$: $f'(x) = \lim_{h \to 0} \frac{-f(h) + 2xh + 1}{h}$. We can split the fraction: $f'(x) = \lim_{h \to 0} \left( \frac{-f(h) + 1}{h} + \frac{2xh}{h} \right)$ $f'(x) = \lim_{h \to 0} \left( -\frac{f(h) - 1}{h} + 2x \right)$.

To evaluate the limit $\lim_{h \to 0} \frac{f(h) - 1}{h}$, we first need to find $f(0)$. Set $x=0$ and $y=0$ in the original functional equation: $f(0+0) = f(0) - f(0) + 2(0)(0) + 1$ $f(0) = f(0) - f(0) + 0 + 1$ $f(0) = 1$.

Now, the limit $\lim_{h \to 0} \frac{f(h) - 1}{h}$ can be written as $\lim_{h \to 0} \frac{f(h) - f(0)}{h}$. By the definition of the derivative at $x=0$, this limit is $f'(0)$. We are given that $f'(0) = 1$. So, $\lim_{h \to 0} \frac{f(h) - 1}{h} = f'(0) = 1$.

Substitute this value back into the expression for $f'(x)$: $f'(x) = -\lim_{h \to 0} \frac{f(h) - 1}{h} + \lim_{h \to 0} 2x$ $f'(x) = -(f'(0)) + 2x$ $f'(x) = -1 + 2x$ $f'(x) = 2x - 1$.

Method 2: Differentiating the functional equation.

Since $f(x)$ is differentiable everywhere, we can differentiate the functional equation with respect to $y$, treating $x$ as a constant. $f(x + y) = f(x) – f(y) + 2xy + 1$ Differentiating both sides with respect to $y$: $\frac{\partial}{\partial y} [f(x + y)] = \frac{\partial}{\partial y} [f(x) – f(y) + 2xy + 1]$ Using the chain rule on the left side, $\frac{\partial}{\partial y} f(x+y) = f'(x+y) \cdot \frac{\partial}{\partial y}(x+y) = f'(x+y) \cdot 1 = f'(x+y)$. On the right side, $\frac{\partial}{\partial y} f(x) = 0$ (since $x$ is treated as a constant). $\frac{\partial}{\partial y} f(y) = f'(y)$. $\frac{\partial}{\partial y} (2xy) = 2x$. $\frac{\partial}{\partial y} (1) = 0$. So, the differentiated equation is: $f'(x+y) = 0 - f'(y) + 2x + 0$ $f'(x+y) = 2x - f'(y)$.

This equation holds for all $x, y \in R$. Let $y=0$: $f'(x+0) = 2x - f'(0)$ $f'(x) = 2x - f'(0)$.

We are given that $f'(0) = 1$. Substitute this value: $f'(x) = 2x - 1$.

Both methods yield the same result.

The function $f'(x)$ is equal to $2x - 1$. This corresponds to option (b).