How do you find the linearization at x=1 of $f(x)= 2x^3-3x$ ?

Question

12840 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-19T11:04:56

The linearization is given by $3x-4$ .

The linearization of a function $f$ at a certain point $x_0$ is the tangent line to $f$ in $x_0$

It is given by $f(x_0)+f'(x_0)(x-x_0)$ .

In your case, $f'(x)=6x^2-3$ , and thus $f'(1)=6-3=3$

Your line is thus

$f(1)+f'(1)(x-1)=-1+3(x-1)=-1+3x-3=3x-4$ .

Here you can see the graphs of the function and the line, checking that it is indeed the tangent in $x=1$ .

How do you find the linearization at x=1 of f(x)= 2x^3-3x f(x)= 2x^3-3x ?