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, $y=14x-4$

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