How to prove that the differential quotient of f with&nbsp;f(x)=x2&nbsp;in the interval&nbsp;[1;&nbsp;b]&nbsp;is equal to the differential quotient of g with&nbsp;g(x)=12(x2+x)&nbsp;in the interval&nbsp;[b;&nbsp;b+1]?

Step 1The differential quotient of a function&nbsp;Φ&nbsp;over&nbsp;[a,&nbsp;b]&nbsp;isΦ(b)−Φ(a)b−a. Note that for f we have:f(b)−f(1)b−1=b2−12b−1=b+1And for g:g(b+1)−g(b)(b+1)−b=12((b+1)2+(b+1))−12(b2+b)=12(b2+2b+1+b+1−b2−b)=12(2b+2)=b+1

Resolved: "How to prove that the differential quotient of..."

High School
Calculus and Analysis
Calculus 2
Differential equations

Camila Glenn

Answered question

2022-04-07

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 $g (x) = \frac{1}{2} (x^{2} + x)$ in the interval $[b; b + 1]$ ?

Answer & Explanation

pobijedi6wro

Beginner2022-04-08Added 15 answers

Step 1
The differential quotient of a function $Φ$ over $[a, b]$ is
$\frac{Φ (b) - Φ (a)}{b - a}$ . Note that for f we have:
$\frac{f (b) - f (1)}{b - 1} = \frac{b^{2} - 1^{2}}{b - 1} = b + 1$
And for g:
$\frac{g (b + 1) - g (b)}{(b + 1) - b}$
$= \frac{1}{2} ({(b + 1)}^{2} + (b + 1)) - \frac{1}{2} (b^{2} + b)$
$= \frac{1}{2} (b^{2} + 2 b + 1 + b + 1 - b^{2} - b)$
$= \frac{1}{2} (2 b + 2)$
$= b + 1$

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