Y=f(x)= 8/sqrt(x-2) (x,y)= (6,4)

EliteGenius

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Y=f(x)= 8/sqrt(x-2) (x,y)= (6,4)

 

Answer No.1

Anonymous

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Y=f(x)= 8/sqrt(x-2) y' =df(x) / dx = -4 *(x-2)^-3/2 at point (6,4) y' =df(x) / dx = -4 *(6-2)^-3/2 y' =df(x) / dx = -4 *(4)^-3/2 = -4/8 = -1/2 eqaution of tangent y -4 = -1/2 (x-6) 2y-8 = -x+6 2y= -x + 14

Answer No.2

Anonymous

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F(x) = 8(x-2)^(-1/2) Use the power rule. f'(x) = 8(-1/2)(x-2)^(-3/2) = -4(x-2)^(-3/2) Find the slope of the tangent line at x = 4. f'(4) = -4(2)^(-3/2) = -sqrt(2) So using the equation of a line. y - 6 = -sqrt(2)(x - 4) y = -sqrt(2)(x - 4) + 6

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