For f(x)=x5−x3+x what is the equation of the tangent line at x=3?

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vefibiongedogn7z · Accepted Answer

The gradient of the tangent line to the curve of f at the point 3 will be equal to the derivative of the function at the point 3.
$∴ f^{'} (x) = 5 x^{4} - 3 x^{2} + 1$
$∴ f^{'} (3) = 5 * 3^{4} - 3 * 3^{2} + 1 = 379$
But at 3, the function itself has value
$f (3) = 3^{5} - 3^{3} + 3 = 219$
Therefore the point with co-ordinates $(x, y) = (3, 219)$ is the point of contact of the tangent with the curve and hence lies on the tangent line so satisfies its equation, so we can substitute it in to yield
$y = m x + c \Rightarrow 219 = (379) (3) + c,$
from which we obtain $c = - 918$ .
Thus the equation of the required tangent line is $y = 379 x - 918$ .

For f(x)=x^{5}-x^{3}+x what is the equation of the tangent line

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