Practicing Success
Suppose that f is a differentiable function with the property that f(x + y) = f(x) + f(y) + xy and $\underset{h→0}{\lim}\frac{1}{h}f(h)=3$, then |
f is a linear function $f(x) = 3x + x^2$ $f(x)=3x+\frac{x^2}{2}$ none of these |
$f(x)=3x+\frac{x^2}{2}$ |
$f'(x)=\underset{h→0}{\lim}\frac{f(x+h)-f(x)}{h}=\underset{h→0}{\lim}\frac{f(x)+f(h)+xh-f(x)}{h}$ $=\underset{h→0}{\lim}\frac{1}{h}f(h)+x=3+x$ Hence $f(x)=3x+\frac{x^2}{2}+c$ Putting x = y = 0 in the given equation, we have $f(0)=f(0+0)=f(0)+f(0)+0⇒f(0)=0$ Thus c = 0 and $f(x)=3x+\frac{x^2}{2}$ |