Camila Glenn

2022-04-07

How to prove that the differential quotient of f with $f\left(x\right)={x}^{2}$ in the interval $[1;\text{}b]$ is equal to the differential quotient of g with $g\left(x\right)=\frac{1}{2}({x}^{2}+x)$ in the interval $[b;\text{}b+1]$?

pobijedi6wro

Beginner2022-04-08Added 15 answers

Step 1

The differential quotient of a function $\mathrm{\Phi}$ over $[a,\text{}b]$ is

$\frac{\mathrm{\Phi}\left(b\right)-\mathrm{\Phi}\left(a\right)}{b-a}$. Note that for f we have:

$\frac{f\left(b\right)-f\left(1\right)}{b-1}=\frac{{b}^{2}-{1}^{2}}{b-1}=b+1$

And for g:

$\frac{g(b+1)-g\left(b\right)}{(b+1)-b}$

$=\frac{1}{2}({(b+1)}^{2}+(b+1))-\frac{1}{2}({b}^{2}+b)$

$=\frac{1}{2}({b}^{2}+2b+1+b+1-{b}^{2}-b)$

$=\frac{1}{2}(2b+2)$

$=b+1$

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