# (a) Find the local linearization of f ( x ) = 1 1 + 3 x near x = 0 . (b) Using your answer to...

## Question:

(a) Find the local linearization of

{eq}f(x) = \frac{1}{1+3x} {/eq}

near {eq}x=0 {/eq}.

(b) Using your answer to (a), what quadratic function would you expect to approximate

{eq}g(x) = \frac{1}{1 + 3x^2} {/eq}?

(c) Using your answer to (b), what would you expect the derivative of {eq}\frac{1}{1 + 3x^2} {/eq} to be even without doing any differentiation?

## Linearization of Functions

Given a radical function we linearize it at a point to get its linear version. Then we are given a second function and we get its quadratic version at a point using an extension of the linearization technique which is simply an application of the Taylor's Theorem from Calculus. The linearization technique is an important tool that helps us understand long-term or asymptotic behavior of solutions of ordinary differential equations as the independent variable goes to infinity.

## Answer and Explanation:

(a) The linearization of {eq}\displaystyle f(x)=\frac {1}{1+3x} {/eq} at {eq}x=0 {/eq} is given by

{eq}L(x)=f(0)+xf'(0) \qquad (1) {/eq}

Now using first derivatives,

{eq}f'(x)=-3(1+3x)^{-2} \implies f'(0) = -3. {/eq}

Substituting all this back in (1) gives us our linearization as

{eq}L(x)=1-3x \qquad (2) {/eq}

(b) From the answer to part (a), the quadratic approximation would be

{eq}Q(x)=1-3x^2 \qquad (3) {/eq}

(c) Since the quadratic approximation is an even polynomial function, the derivative of the function would be an odd polynomial function and therefore NOT even.

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