What is the equation of the line tangent to #f(x)=2x ^3+8x # at #x=1#?

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Answer 1 · 2016-04-01T09:30:48

#y=14x-4#

The equation of the tangent line is: #y=kx+n#
where #k=f'(x_0), x_0=1#

So, #f'(x)=(2x^3+8x)'=6x^2+8#,

and hence #k=f'(1)=6*1^2+8=14#

We have: #f(x_0)=f(1)=2*1^3+8*1=10# and

#y(x_0)=kx_0+n#

#10=14*1+n => n=-4#

Finally, #y=14x-4#