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

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Answer 1 · 2016-10-05T16:33:53

#y=-x-2#

Find the derivative of the function.

Begin by using the Product Rule.

#f'(x)=uv'+u'v#

#u=x^3-3x+1#
#u'=3x^2-3#

#v=x+2#
#v'=1#

#f'(x)=(x^3-3x+1)(1)+(3x^2-3)(x+2)#

#f'(x)=(x^3-3x+1)+(3x^2-3)(x+2)#

Substitute in the value of #x# to get a numerical value for the slope, #m#.

#f'(1)=((1)^3-3(1)+1)+(3(1)^2-3)((1)+2)#

Simplify

#f'(1)=(1-3+1)+(3-3)(1+2)#

#f'(1)=-1+(0)(3)=-1+0=-1#

The slope, #m#, is #-1#.

Substitute in the point #(1,-3)# and the slope, #m#, #-1#, into the slope intercept formula, #y=mx+b#.

#-3=-1(1)+b#

Simplify

#-3=-1+b#

Add #1# to both sides of the equation to isolate #b#.

#-2=b#

Write the equation of the tangent line using the above information.

#y=-1x-2#

#or#

#y=-x-2#

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