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Question Number 1133 by 112358 last updated on 29/Jun/15

$$Letf:[0,1]\to \mathbb{R}bea$$$$differentiablefunction.Prove$$$$thatthereexistsac\in [0,1]such$$$$that$$$$\text{Prime causes double exponent: use braces to clarify}$$

Commented by 123456 last updated on 29/Jun/15

$$\mathrm{if}f\mathrm{is}\mathrm{diferrentiable}\mathrm{into}[a,b]\mathrm{then}$$$$\mathrm{\exists }\xi \in [a,b]\Rightarrow f^{\prime }\left(\xi \right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}(b>a)$$

Commented by 123456 last updated on 29/Jun/15

$$f\mathrm{is}\mathrm{continuous}\mathrm{at}[a,b]\mathrm{then}$$$$m=\min \{f\left(a\right),f\left(b\right)\}$$$$M=\max \{f\left(a\right),f\left(b\right)\}$$$$c\in [m,\mathrm{M}]\Rightarrow \mathrm{\exists }\xi \in [a,b],f\left(\xi \right)=c$$

Commented by 123456 last updated on 30/Jun/15

$$f^{\prime }\left(c\right)=\frac{4[f\left(1\right)-f\left(0\right)]}{\pi (1+c^2)}$$$$f\left(c\right)=\frac{4[f\left(1\right)-f\left(0\right)]}{\pi }\arctan c$$

Commented by 123456 last updated on 01/Jul/15

$$f\left(x\right)=ax+b$$$$f\left(1\right)=a+b$$$$f\left(0\right)=b$$$$f\left(1\right)-f\left(0\right)=a$$$$\frac{4[f\left(1\right)-f\left(0\right)]}{\pi }=\frac{4a}{\pi }$$$$f^{\prime }\left(x\right)=a$$$$(1+c^2)f^{\prime }\left(c\right)=(1+c^2)a$$$$c^2+1=\frac{4}{\pi }$$$$c=\pm \sqrt{\frac{4}{\pi }-1}$$

Commented by 123456 last updated on 03/Jul/15

$$g\left(x\right)=\arctan x$$

Answered by 123456 last updated on 03/Jul/15

$$g\left(x\right)=\arctan x$$$$\frac{f\left(1\right)-f\left(0\right)}{g\left(1\right)-g\left(0\right)}=\frac{f^{\prime }\left(c\right)}{g^{\prime }\left(c\right)}$$

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