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I was hoping for some help with an old question of mine (got it wrong on homework a while ago- was sick the day this material was taught- never quite got the question).

Find the linearization L(x) off(x)=3x^4-5x^3 at x=2

Hello Clark,

I gather you are taking calculus, since this is a standard calculus question.

The linearization is just the tangent line approximation at the point given.

So, we just need the equation of the tangent line to f(x) at x=2.

First, let's find the y-value at x=2 ==> y=f(2)=8, so the point of tangency is (2,8).

Now for the slope, f'(2). f'(x)=12x^3-15x^2 ==> f'(2)=36.

The line with slope 36 though (2,8) ==> y-8=36(x-2) ==> y-8=36x-72 ==> y=36x-64.

Thus, L(x)=36x-64

A. Mantell

Calculus

Answers by Expert:

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

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