# Approximate the root by using a linearization centered at an appropriate nearby number. \sqrt{147}

## Question:

Approximate the root by using a linearization centered at an appropriate nearby number.

{eq}\sqrt{147} {/eq}

## Linear Approximation:

A strategy which is estimating a function at an immediate point is called linear approximation or linearization.

Now:

If we speculate that the function is {eq}f(x) {/eq} and the near by point is {eq}x = a {/eq}

Then:

Linearization can be clarified by the subsequent mathematical formula:

{eq}L(x) = f(a) + f'(a) (x-a) {/eq}

## Answer and Explanation:

Let us assume that:

{eq}f(x) = \sqrt{x} {/eq}

And:

The nearest point is:

{eq}a = 144 {/eq}

Derivative of {eq}f(x) {/eq} with respect to {eq}x {/eq} is:

{eq}\begin{align*} f^{'} (x) &= \frac{d}{dx} (x^{\frac{1}{2}}) \\ &= \frac{1}{2} x^{(\frac{1}{2} - 1)} \hspace{1 cm} \left[ \because \frac{d}{dx} (x^n) = nx^{n-1} \right] \\ &= \frac{1}{2} x^{ - \frac{1}{2}} \\ &= \frac{1}{2 x^{\frac{1}{2}}} \\ &= \frac{1}{2 \sqrt{x}} \end{align*} {/eq}

Now:

Using linerization, we can write:

{eq}\begin{align*} L(147) &= f(144) + (147 - 144) f^{'} (144) \hspace{1 cm} \left[ \because L( x) = f(a) + f^{'} (a) (x-a) \right] \\ &= \sqrt{144} + \frac{3}{2 \sqrt{144}} \hspace{1 cm} \left[ \because f(x) = \sqrt{x} \, and \, f^{'} (x) = \frac{1}{2 \sqrt{x}} \right] \\ &= 12 + \frac{3}{24} \\ &= 12 + \frac{1}{8} \\ &= 12.125 \end{align*} {/eq}

Hence:

Using linerization, the value of {eq}\sqrt{147} {/eq} is {eq}12.125 {/eq}

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#### Learn more about this topic:

from Math 104: Calculus

Chapter 10 / Lesson 2