Answer
The equation of the tangent is $y=-3x-1$.
Work Step by Step
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x^3+4x-1 ;u’(x)=3x^2+4 $.
$v(x)=x-2 ;v’(x)=1 $.
$f'(x)=(3x^2+4)(x-2)+(x^3+4x-1)$.
$f'(1)=(3(1^2)+4)(1-2)+(1^3+4(1)-1)=-3$
Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y+4)=-3x+3 \rightarrow y=-3x-1$
A graphing calculator and a computer algebra system have been used to confirm these results.
