What is the equation of the tangent line of #f(x)=x^3-2x-3# at #x=-1#?

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Answer 1 · 2018-04-11T05:30:49

#y=x-1#

Given: #f(x)=x^3-2x-3#

We first need to find the slope of the tangent line at #x=-1#, and so we need to differentiate the function.

#f'(x)=3x^2-2#

#=>f'(-1)=3*(-1)^2-2#

#=3*1-2#

#=1#

Then, we need to find what is #f(x)# at #x=-1#, i.e. #f(-1)#.

#f(-1)=(-1)^3-2*-1-3#

#=-1+2-3#

#=-4+2#

#=-2#

So, the point is at #(-1,-2)#.

Now, we use the point-slope form to find the equation.

#y-y_0=m(x-x_0)#

#y-(-2)=1(x-(-1)#

#y+2=x+1#

#y=x+1-2#

#=x-1#

A graph proves it:

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