Differential quotient problem
How to prove that the differential quotient of $f$ with $f(x)=x^2$ in the interval $[1;b]$ is equal to the differential quotient of $g$ with $g(x)=\dfrac 1 2(x^2+x)$ in the interval $[b;b+1]$?
The differential quotient of a function $\Phi$ over $[a,b]$ is $\dfrac{\Phi(b)-\Phi(a)}{b-a}$. Note that for $f$ we have:
$\dfrac{f(b)-f(1)}{b-1}=\dfrac{b^2-1^2}{b-1}=b+1$.
And for $g$:
$\begin{array}{cl} \dfrac{g(b+1)-g(b)}{(b+1)-b} & =\frac{1}{2}\left((b+1)^2+(b+1)\right)-\frac{1}{2}(b^2+b)\\ & =\frac{1}{2}(b^2+2b+1+b+1-b^2-b)\\ & =\frac{1}{2}(2b+2)\\ & =b+1\end{array}$