Question

Suppose that $f$ is a differentiable function with the property that $f(x+y)=f\left(x\right)+f\left(y\right)+xy$ and $\underset{h\to 0}{lim}\frac{1}{h}f\left(h\right)=3$, then

• A. $f$ is a linear function
• B. $f\left(x\right)=3x+x^2$
• C. $f\left(x\right)=3x+\frac{x^2}{2}$
• D. $f\left(x\right)=3x-\frac{x^2}{2}$

Answer 1

The correct option is C $f\left(x\right)=3x+\frac{x^2}{2}$

$f(x+h)=f\left(x\right)+f\left(h\right)+xh$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}+x$

$\Rightarrow f^{\prime }\left(x\right)=3+x$

$f\left(x\right)=3x+\frac{x^2}{2}+cf\left(0\right)=0$

$\therefore c=0$

$f\left(x\right)=3x+\frac{x^2}{2}$