For #f(x)=x^5-x^3+x# what is the equation of the tangent line at #x=3#?

1 Answer
Oct 30, 2015

#y=379x-918#

Explanation:

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.

#thereforef'(x)=5x^4-3x^2+1#

#thereforef'(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=mx+c=>219=(379)(3)+c#,
from which we obtain #c=-918#.

Thus the equation of the required tangent line is #y=379x-918#.