find the equation of the tangent and the equation of the normal to the curve y = x - 1/3 x^3 + 1/7 x^5 where x = 4

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The tangent line to the given curve, at the point x = 4 is the derivative of the function at x = 4.

Therefore, we need to determine the equation of the 1st derivative of the function.

If the function is y = x - 1/(3x^3) + 1/(7x^5), we'll get:

y' = 1 - (-9x^2)/9x^6 + (-35x^4)/49x^10

y' = 1 + 1/x^4 - 5/7x^6

We'll calculate the slope of the tangent line, at the point x = 4:

y'(4) = 1 + 1/256 - 5/7*4096

y'(4) = (28672 +112 - 5 )

y'(4) = 28779/28672

We'll calculate the value of the function at x = 4:

y(4) = 4 - 1/192 + 1/7168

y(4) = (4*7168 - 112 + 3)/21504

y(4) = 28563/21504

Therefore, the equation of the tangent line at the curve y = x - 1/(3x^3) + 1/(7x^5) t x = 4, is y - 28563/21504 = 28779(x-4)/28672.

The equation of the normal line is written: y - 28563/21504 = -28672(x-4)/28779.

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