(a) find the local linearization of\n\n f(x)=\frac{1}{1+5 x} \n\nnear ( x=0 ):\n\n \frac{1}{1+5 x} approx…

(a) find the local linearization of\n\n f(x)=\frac{1}{1+5 x} \n\nnear ( x=0 ):\n\n \frac{1}{1+5 x} approx \n\n(b) using your answer to (a), what quadratic function would you expect to approximate\n\n g(x)=\frac{1}{1+5 x^{2}} \n\n \frac{1}{1+5 x^{2}} approx \n\n(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?\n\n left.\frac{d}{d x}left(\frac{1}{1+5 x^{2}}\right)\right|_{x=0}= \n

(a) find the local linearization of\n\n f(x)=\frac{1}{1+5 x} \n\nnear ( x=0 ):\n\n \frac{1}{1+5 x} approx \n\n(b) using your answer to (a), what quadratic function would you expect to approximate\n\n g(x)=\frac{1}{1+5 x^{2}} \n\n \frac{1}{1+5 x^{2}} approx \n\n(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?\n\n left.\frac{d}{d x}left(\frac{1}{1+5 x^{2}}\right)\right|_{x=0}= \n

Answer

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)