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Question Number 183158 by cortano1 last updated on 21/Dec/22

Given f(x)= (([(1/3)x]∣2x∣+Ax)/(∣4−x^2 ∣)) if f ′(−1)= 5 then A=? [ ] = floor function

$$\:{Given}\:{f}\left({x}\right)=\:\frac{\left[\frac{\mathrm{1}}{\mathrm{3}}{x}\right]\mid\mathrm{2}{x}\mid+{Ax}}{\mid\mathrm{4}−{x}^{\mathrm{2}} \mid} \\ $$$$\:{if}\:{f}\:'\left(−\mathrm{1}\right)=\:\mathrm{5}\:{then}\:{A}=? \\ $$$$\left[\:\:\:\:\right]\:=\:{floor}\:{function}\: \\ $$

Answered by TheSupreme last updated on 21/Dec/22

f(x)=((g(x)h(x))/(u(x)))+((Ax)/(4−x^2 )) f′(−1)=(((g′(x)h(x)+g(x)h′(x))u(x)−g(x)h(x)u′(x))/(u^2 (x)))+A ((4−x^2 −2x^2 )/((4−x^2 )^2 ))= g′(−1)=0 g(−1)=−1 h′(−1)=2 h(−1)=2 u′(−1)=6 u(−1)=3 f′(x)=(((−2+2)3+2∗6)/3^2 )+A (1/9)=5 12+A=45 → A=32

$${f}\left({x}\right)=\frac{{g}\left({x}\right){h}\left({x}\right)}{{u}\left({x}\right)}+\frac{{Ax}}{\mathrm{4}−{x}^{\mathrm{2}} } \\ $$$${f}'\left(−\mathrm{1}\right)=\frac{\left({g}'\left({x}\right){h}\left({x}\right)+{g}\left({x}\right){h}'\left({x}\right)\right){u}\left({x}\right)−{g}\left({x}\right){h}\left({x}\right){u}'\left({x}\right)}{{u}^{\mathrm{2}} \left({x}\right)}+{A}\:\frac{\mathrm{4}−{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{4}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }= \\ $$$${g}'\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$${g}\left(−\mathrm{1}\right)=−\mathrm{1} \\ $$$${h}'\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${h}\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${u}'\left(−\mathrm{1}\right)=\mathrm{6} \\ $$$${u}\left(−\mathrm{1}\right)=\mathrm{3} \\ $$$${f}'\left({x}\right)=\frac{\left(−\mathrm{2}+\mathrm{2}\right)\mathrm{3}+\mathrm{2}\ast\mathrm{6}}{\mathrm{3}^{\mathrm{2}} }+{A}\:\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{5} \\ $$$$\mathrm{12}+{A}=\mathrm{45}\:\rightarrow\:{A}=\mathrm{32} \\ $$

Commented by greougoury555 last updated on 22/Dec/22

what the formula ((d[(1/3)x])/dx) =?

$${what}\:{the}\:{formula}\:\frac{{d}\left[\frac{\mathrm{1}}{\mathrm{3}}{x}\right]}{{dx}}\:=? \\ $$

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