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Question Number 215760 by MATHEMATICSAM last updated on 17/Jan/25

$$\mathrm{Let}g\left(x\right)=3f\left(\frac{x}{3}\right)+f(3-x)\mathrm{and}$$$$f{}^{″}\left(x\right)>0\mathrm{for}\mathrm{all}x\in (0,3).\mathrm{If}g\mathrm{is}$$$$\mathrm{decreasing}\mathrm{in}(0,\alpha )\mathrm{and}\mathrm{increasing}\mathrm{in}$$$$(\alpha ,3)\mathrm{then}\mathrm{find}8\alpha .$$

Answered by universe last updated on 18/Jan/25

$$f^{″}\left(x\right)>0=>f^{\prime }isstrictlyincreasing$$$$g^{\prime }\left(x\right)=f^{\prime }\left(x/3\right)-f^{\prime }(3-x)$$$$g^{\prime }\left(x\right)=0$$$$\frac{x}{3}=3-x$$$$x+x/3=3$$$$x=\frac{9}{4}isacriticalpoint$$$$g\left(x\right)isdecreing(0,9/4)$$$$\alpha =9/4$$$$8\alpha =18$$$$$$

Commented by MATHEMATICSAM last updated on 17/Jan/25

$$\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{given}18$$

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