# Find the linearization L(x) of the function at a. f(x)=x^4-+3x^2, a=-1Z

Find the linearization L(x) of the function at a. $$\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}\mp{3}{x}^{{2}}$$, $$\displaystyle{a}=-{1}$$

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wheezym

Linearization through tangent line approximation is achieved with equation:
$$\displaystyle{L}{\left({x}\right)}\approx{f{{\left({a}\right)}}}+{f}{\left({a}\right)}\cdot{\left({x}-{a}\right)}$$
The derivative of f(x) is:
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}+{3}{x}^{{{2}}}$$
$$\displaystyle{f}{\left({x}\right)}={\left({x}^{{{4}}}+{3}{x}^{{{2}}}\right)}$$
$$\displaystyle={\left({x}^{{{4}}}\right)}+{\left({3}{x}^{{{2}}}\right)}$$
$$\displaystyle={4}{x}^{{{3}}}+{2}\cdot{3}{x}^{{{1}}}$$
$$\displaystyle={4}{x}^{{{3}}}+{6}{x}$$
And with $$\displaystyle{f{{\left({a}=-{1}\right)}}}={4}$$ and $$f(a=-1)=-10$$ our linearization model is:
$$\displaystyle{L}{\left({x}\right)}={4}-{10}\cdot{\left({x}+{1}\right)}$$
$$\displaystyle={4}-{10}\cdot{x}-{10}$$
$$\displaystyle{L}{\left({x}\right)}=-{6}-{10}\cdot{x}$$