A function is shown. What is the value of f(–8)?

Mathematics

1 answer:

Step2247 [10]2 years ago

8 0

Answer:36

Step-by-step explanation:

f(x)=12-3*(-8)

=36

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Find the derivatives of the following implicit function

ddd [48]

Answer:

$\frac{d}{dx}\left(y\right)=\frac{2-6x+6y}{-6x+2y+1}$

Step-by-step explanation:

$3x^2-6xy+y^2=2x-y\\\mathrm{Treat\:}y\mathrm{\:as\:}y\left(x\right)\\\mathrm{Differentiate\:both\:sides\:of\:the\:equation\:with\:respect\:to\:}x\\\frac{d}{dx}\left(3x^2-6xy+y^2\right)=\frac{d}{dx}\left(2x-y\right)\\\frac{d}{dx}\left(3x^2-6xy+y^2\right)=6x-6\left(y+x\frac{d}{dx}\left(y\right)\right)+2y\frac{d}{dx}\left(y\right)\\\frac{d}{dx}\left(2x-y\right)=2-\frac{d}{dx}\left(y\right)\\6x-6\left(y+x\frac{d}{dx}\left(y\right)\right)+2y\frac{d}{dx}\left(y\right)=2-\frac{d}{dx}\left(y\right)$

$\mathrm{For\:convenience,\:write\:}\frac{d}{dx}\left(y\right)\mathrm{\:as\:}y^{'\:}\\6x-6\left(y+xy^{'\:}\right)+2yy^{'\:}=2-y^{'\:}\\\mathrm{Isolate}\:y^{'\:}:\quad y^{'\:}=\frac{2-6x+6y}{-6x+2y+1}\\y^{'\:}=\frac{2-6x+6y}{-6x+2y+1}\\\mathrm{Write}\:y^{'\:}\:\mathrm{as}\:\frac{d}{dx}\left(y\right)\\\frac{d}{dx}\left(y\right)=\frac{2-6x+6y}{-6x+2y+1}$