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Given-f-x-1-3-x-2x-Ax-4-x-2-if-f-1-5-then-A-floor-function-

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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
Givenf(x)=[13x]∣2x∣+Ax∣4−x2∣iff′(−1)=5thenA=?[]=floorfunction
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(x)=g(x)h(x)u(x)+Ax4−x2f′(−1)=(g′(x)h(x)+g(x)h′(x))u(x)−g(x)h(x)u′(x)u2(x)+A4−x2−2x2(4−x2)2=g′(−1)=0g(−1)=−1h′(−1)=2h(−1)=2u′(−1)=6u(−1)=3f′(x)=(−2+2)3+2∗632+A19=512+A=45→A=32
Commented by greougoury555 last updated on 22/Dec/22
what the formula ((d[(1/3)x])/dx) =?
whattheformulad[13x]dx=?

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