Calculus (3rd Edition)

$\delta f(x)=2x+1$ $\delta f(x)=1$ $\delta f(x)=3x^2+3x+1$
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: $\delta f(x)=3x^2+3x+1$