How do you find the linearization at a=1 of $f (x) = x^{4} + 4 x^{2}$ $f(x) = x^4 + 4x^2$ ?

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How do you find the linearization at a=1 of $f (x) = x^{4} + 4 x^{2}$ $f(x) = x^4 + 4x^2$ ?

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Answer 1 · 2015-11-05T20:26:47

A linearization is a tangent line.

Explanation:

The linearization of $f (x) = x^{4} + 4 x^{2}$ $f(x) = x^4 + 4x^2$ at $a = 1$ $a=1$ is one form of the equation of the line tangent to the graph of $f$ $f$ at the point $(1, f (1))$ $(1,f(1))$ .

For $f (x) = x^{4} + 4 x^{2}$ $f(x) = x^4 + 4x^2$ , we have $f (1) = 5$ $f(1) = 5$ and $f'(x) = 4x^3+8x$ , so $f'(1) = 12$ .

The tangent line has point-slope form:

$y-f(1) = f'(1)(x-1)$ $" "$ $" "$ $y-5=12(x-1)$ .

The linearization is:

$y=f(1) + f'(1)(x-1)$ $" "$ $" "$ $y=5 + 12(x-1)$ .

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How do you find the linearization at a=1 of $f (x) = x^{4} + 4 x^{2}$ $f(x) = x^4 + 4x^2$ ?

1 Answer

Explanation:

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How do you find the linearization at a=1 of f(x) = x^4 + 4x^2f(x)=x4+4x2 f(x) = x^4 + 4x^2?

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How do you find the linearization at a=1 of $f (x) = x^{4} + 4 x^{2}$ $f(x) = x^4 + 4x^2$ ?