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Question Number 146523 by mnjuly1970 last updated on 13/Jul/21

Answered by ajfour last updated on 13/Jul/21

$$f\left(\frac{x+y}{2}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$$$$f\left(0\right)=2$$$$f\left(x\right)=2x+2$$$$3f(x+y)=2+f\left(2x\right)+f\left(2y\right)$$$$\Rightarrow$$$$3[2(x+y)+2]=2+4x+2$$$$+4y+2$$$$\Rightarrow x+y=0$$$$soy=-x\Rightarrow$$$$3f\left(0\right)=2+f\left(2x\right)+f(-2x)$$$$butf\left(0\right)=2\Rightarrow$$$$f\left(2x\right)+f(-2x)=4$$$$2f{}^{\prime }\left(2x\right)-2f{}^{\prime }(-2x)=0$$$$\Rightarrow f{}^{\prime }\left(2x\right)=f{}^{\prime }(-2x)$$$$g\left(x\right)=\mid f(\mid x\mid )-3\mid$$$$=\mid 2\mid x\mid +2-3\mid$$$$=\mid 2\mid x\mid -1\mid$$$$g\left(x\right){isn}^{\prime }tdifferentiableat$$$$x=\pm \frac{1}{2},x=0$$

Commented by ajfour last updated on 13/Jul/21

Commented by mnjuly1970 last updated on 14/Jul/21

$$thankyousomuch...$$

Answered by mathmax by abdo last updated on 13/Jul/21

$$\mathrm{y}=\mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{2+2\mathrm{f}\left(\mathrm{x}\right)}{3}\Rightarrow 3\mathrm{f}\left(\mathrm{x}\right)=2+2\mathrm{f}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=2\mathrm{how}\mathrm{to}\mathrm{get}$$$${\mathrm{f}}^{{}^{\prime }}\left(2\right)=2\mathrm{if}\mathrm{f}\mathrm{is}\mathrm{constant}?\mathrm{something}\mathrm{went}\mathrm{wrong}....$$

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