Question

(a) find the local linearization of

f(x)=\frac{1}{1+5 x}

near ( x=0 ):

\frac{1}{1+5 x} approx

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

g(x)=\frac{1}{1+5 x^{2}}

\frac{1}{1+5 x^{2}} approx

(c) using your answer to (b), what would you expect the derivative of ( \frac{1}{1+5 x^{2}} ) to be even without doing any differentiation?

left.\frac{d}{d x}left(\frac{1}{1+5 x^{2}}
ight)
ight|_{x=0}=

Explanation:

Step1: Recall the local linearization formula

The local linearization of \(y = f(x)\) at \(x = a\) is \(L(x)=f(a)+f^{\prime}(a)(x - a)\). For \(f(x)=\frac{1}{1 + 5x}=(1 + 5x)^{-1}\), \(a = 0\). First, find \(f(0)\): \(f(0)=\frac{1}{1+5\times0}=1\). Then, find the derivative using the chain - rule. If \(u = 1 + 5x\), \(y = u^{-1}\), \(\frac{dy}{du}=-u^{-2}\) and \(\frac{du}{dx}=5\). So \(f^{\prime}(x)=-5(1 + 5x)^{-2}\). Then \(f^{\prime}(0)=-5\).
The local linearization \(L(x)=f(0)+f^{\prime}(0)(x - 0)=1-5x\). So \(\frac{1}{1 + 5x}\approx1-5x\).

Step2: Use substitution for the quadratic approximation

If \(\frac{1}{1 + 5x}\approx1-5x\), then for \(g(x)=\frac{1}{1 + 5x^{2}}\), substitute \(x\) with \(x^{2}\). So \(\frac{1}{1 + 5x^{2}}\approx1-5x^{2}\).

Step3: Differentiate the quadratic approximation

If \(h(x)=1-5x^{2}\), then \(h^{\prime}(x)=-10x\). Evaluate at \(x = 0\), \(h^{\prime}(0)=0\).

Answer:

(a) \(1-5x\)
(b) \(1 - 5x^{2}\)
(c) \(0\)