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Question Number 66279 by gunawan last updated on 12/Aug/19

$$\mathrm{Value}\mathrm{of}\mathrm{maximum}$$$$f\left(x\right){=}^2\log (x+5){+}^2\log (3-x)$$$$\mathrm{is}...$$$$\mathrm{a}.4$$$$\mathrm{b}.8$$$$\mathrm{c}.12$$$$\mathrm{d}.15$$$$\mathrm{e}.16$$

Commented by mathmax by abdo last updated on 12/Aug/19

$$f\left(x\right)={log}_2(x+5)+{log}_2(3-x)$$$$x\in D_f\iff x+5>0and3-x>0\iff -5<x<3\Rightarrow D_f=]-5,3[$$$$f\left(x\right)=\frac{1}{ln\left(2\right)}\{ln(x+5)+ln(3-x)\}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{1}{ln2}\{\frac{1}{x+5}-\frac{1}{3-x}\}$$$$=\frac{1}{ln2}\left\{\frac{3-x-x-5}{(x+5)(3-x)}\right\}=\frac{1}{ln\left(2\right)}\left\{\frac{-2-2x}{(x+5)(3-x)}\right\}$$$$=\frac{-2}{ln2}\left\{\frac{x+1}{(x+5)(3-x)}\right\}so{f}^{{}^{\prime }}\left(x\right)=0\iff x=-1$$$$variationoff\left(x\right)$$$$x-5-13$$$$f^{{}^{\prime }}\left(x\right)+0-$$$$f\left(x\right)-\mathrm{\infty }incf(-1)decr-\mathrm{\infty }$$$$maxf\left(x\right)=f(-1)=\frac{1}{ln\left(2\right)}\{2ln\left(2\right)+2ln\left(2\right)\}=4$$$$x\in D_f$$$$thecorrectanswerisa)$$

Answered by Kunal12588 last updated on 12/Aug/19

$$y={log}_2(x+5)+{log}_2(3-x)$$$$\Rightarrow y=\frac{1}{ln2}(ln(x+5)+ln(3-x))$$$$\Rightarrow y^{\prime }=\frac{1}{ln2}(\frac{1}{x+5}+\frac{-1}{3-x})=\frac{3-x-x-5}{ln2(x+5)(3-x)}$$$$\Rightarrow y^{\prime }=\frac{-2-2x}{ln2(x+5)(3-x)}$$$$formaxormin$$$$y^{\prime }=\frac{-2-2x}{ln2(x+5)(3-x)}=0\Rightarrow (-2-2x)=0$$$$\Rightarrow x=-1$$$$y^{″}=\frac{(x+5)(3-x)(-2)-(-2-2x)\frac{d}{dx}(x+5)(3-x)}{ln2{(x+5)}^2{(3-x)}^2}$$$$puttingx=-1$$$$\Rightarrow y^{″}=\frac{\left(4\right)\left(4\right)(-2)-0}{ln2{(x+5)}^2{(3-x)}^2}<0$$$$\therefore x=-1ispointofmaxima$$$$\therefore maxoff\left(x\right)=f(-1)={log}_2\left(4\right)+{log}_2\left(4\right)=4$$

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