Proof of Mean Value Theorem with Rolle's Theorem

Theorem: if ff is continuous on [a,b][a,b] and ff is differentiable on (a,b)(a,b), then f(b)f(a)ba=f(c)\frac{f(b)-f(a)}{b-a}=f^\prime(c) for some cc in (a,b)(a,b).

Rolle's Theorem: if gg is continuous on [a,b][a,b], gg is differentiable on (a,b)(a,b) and g(b)=g(a)g(b)=g(a), then there is cc in (a,b)(a,b) such that g(c)=0g^\prime(c)=0.

Proof:

Let g(x)=f(x)(f(b)f(a)ba)x\text{Let }g(x)=f(x)-\left(\frac{f(b)-f(a)}{b-a}\right)x

g(a)=f(a)(f(b)f(a)ba)a=f(a)(aba)f(b)+(aba)f(a)=(1+aba)f(a)(aba)f(b)=(bba)f(a)(aba)f(b)\begin{aligned} g(a)&=f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)a\\ &=f(a)-\left(\frac{a}{b-a}\right)f(b)+\left(\frac{a}{b-a}\right)f(a)\\ &=\left(1+\frac{a}{b-a}\right)f(a)-\left(\frac{a}{b-a}\right)f(b)\\ &=\left(\frac{b}{b-a}\right)f(a)-\left(\frac{a}{b-a}\right)f(b) \end{aligned}

g(b)=f(b)(f(b)f(a)ba)b=f(b)(bba)f(b)+(bba)f(a)=(1bba)f(b)+(bba)f(a)=(aab)f(b)+(bba)f(a)\begin{aligned} g(b)&=f(b)-\left(\frac{f(b)-f(a)}{b-a}\right)b\\ &=f(b)-\left(\frac{b}{b-a}\right)f(b)+\left(\frac{b}{b-a}\right)f(a)\\ &=\left(1-\frac{b}{b-a}\right)f(b)+\left(\frac{b}{b-a}\right)f(a)\\ &=-\left(\frac{a}{a-b}\right)f(b)+\left(\frac{b}{b-a}\right)f(a) \end{aligned}

g(b)=g(a)g(c)=0g(b)=g(a)\Rightarrow g^\prime(c)=0

g(x)=f(x)(f(b)f(a)ba)xg(x)=f(x)f(b)f(a)bag(c)=f(c)f(b)f(a)ba0=f(c)f(b)f(a)baf(b)f(a)ba=f(c)\begin{aligned} g(x)&=f(x)-\left(\frac{f(b)-f(a)}{b-a}\right)x\\ g^\prime(x)&=f^\prime(x)-\frac{f(b)-f(a)}{b-a}\\ g^\prime(c)&=f^\prime(c)-\frac{f(b)-f(a)}{b-a}\\ 0&=f^\prime(c)-\frac{f(b)-f(a)}{b-a}\\ \frac{f(b)-f(a)}{b-a}&=f^\prime(c)\quad\blacksquare \end{aligned}

Note by Gordon Chan
1 week, 3 days ago

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