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

$$Givenf\left(x\right)=\frac{\left[\frac{1}{3}x\right]\mid 2x\mid +Ax}{\mid 4-x^2\mid }$$$$iff{}^{\prime }(-1)=5thenA=?$$$$\left[\right]=floorfunction$$

Answered by TheSupreme last updated on 21/Dec/22

$$f\left(x\right)=\frac{g\left(x\right)h\left(x\right)}{u\left(x\right)}+\frac{Ax}{4-x^2}$$$$f^{\prime }(-1)=\frac{(g^{\prime }\left(x\right)h\left(x\right)+g\left(x\right)h^{\prime }\left(x\right))u\left(x\right)-g\left(x\right)h\left(x\right)u^{\prime }\left(x\right)}{u^2\left(x\right)}+A\frac{4-x^2-2x^2}{{(4-x^2)}^2}=$$$$g^{\prime }(-1)=0$$$$g(-1)=-1$$$$h^{\prime }(-1)=2$$$$h(-1)=2$$$$u^{\prime }(-1)=6$$$$u(-1)=3$$$$f^{\prime }\left(x\right)=\frac{(-2+2)3+2\ast 6}{3^2}+A\frac{1}{9}=5$$$$12+A=45\to A=32$$

Commented by greougoury555 last updated on 22/Dec/22

$$whattheformula\frac{d\left[\frac{1}{3}x\right]}{dx}=?$$

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