Answer
$\delta f(x)=2x+1$
$\delta f(x)=1$
$\delta f(x)=3x^2+3x+1$
Work Step by Step
We are given the function:
$f(x)=x^2$
Compute $\delta f(x)$:
$\delta f(x)=f(x+1)-f(x)=(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$
We got:
$\delta f(x)=2x+1$
We are given the function:
$f(x)=x$
Compute $\delta f(x)$:
$\delta f(x)=f(x+1)-f(x)=(x+1)-x=x+1-x=1$
We got:
$\delta f(x)=1$
We are given the function:
$f(x)=x^3$
Compute $\delta f(x)$:
$\delta f(x)=f(x+1)-f(x)=(x+1)^3-x^3$
$=x^3+3x^2+3x+1-x^3$
$=3x^2+3x+1$
We got:
$\delta f(x)=3x^2+3x+1$
