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  • For the curve y = 4x raised to 3 – 2x raised to 5, find all the points at which the tangent passes through the origin.

18) For the curve y = 4x ^ 3 - 2x ^5, find all the points at which the tangent passes
through the origin.

Answers (1)

best_answer

Tangent passes through origin so, (x,y) = (0,0)
Given equation of curve is y = 4x ^ 3 - 2x ^5
Slope of tangent =

\frac{dy}{dx} = 12x^2 - 10x^4
Now, the equation of tangent is 
Y-y= m(X-x)
at (0,0)  Y =  0 and X = 0
-y= (12x^3-10x^4)(-x)
y= 12x^3-10x^5
and we have y = 4x ^ 3 - 2x ^5
4x^3-2x^5= 12x^3-10x^5
8x^5 - 8x^3=0\\ 8x^3(x^2-1)=0\\ x=0\ \ \ \ \ \ and \ \ \ \ \ \ \ x = \pm1
Now, when x = 0,

  y = 4(0) ^ 3 - 2(0) ^5 = 0
when x = 1 , 

y = 4(1) ^ 3 - 2(1) ^5 = 4-2=2
when x= -1 ,

 y = 4(-1) ^ 3 - 2(-1) ^5 = -4-(-2)=-4+2=-2
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

Posted by

Gautam harsolia

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