# How do you find a linear approximation to root(4)(84) ?

Feb 6, 2018

$\sqrt{84} \approx 3.03$

#### Explanation:

Note that ${3}^{4} = 81$, which is close to $84$.

So $\sqrt{84}$ is a little larger than $3$.

To get a better approximation, we can use a linear approximation, a.k.a. Newton's method.

Define:

$f \left(x\right) = {x}^{4} - 84$

Then:

$f ' \left(x\right) = 4 {x}^{3}$

and given an approximate zero $x = a$ of $f \left(x\right)$, a better approximation is:

$a - \frac{f \left(a\right)}{f ' \left(a\right)}$

So in our case, putting $a = 3$, a better approximation is:

$3 - \frac{f \left(3\right)}{f ' \left(3\right)} = 3 - \frac{{3}^{4} - 84}{4 {\left(3\right)}^{3}} = 3 - \frac{81 - 84}{4 \cdot 27} = 3 + \frac{1}{36} = \frac{109}{36} = 3.02 \overline{7}$

This is almost accurate to $4$ significant figures, but let's quote the approximation as $3.03$

Feb 6, 2018

$\sqrt{84} \approx 3.02778$

#### Explanation:

Note that the linear approximation near a point $a$ can be given by:

$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

If given: $f \left(x\right) = \sqrt{x}$

then a suitable choice for $a$ would be $a = 81$ because we know $\sqrt{81} = 3$ exactly and it is close to $84$.

So:

$f \left(a\right) = f \left(81\right) = \sqrt{81} = 3$

Also;

$f \left(x\right) = {x}^{\frac{1}{4}}$ so $f ' \left(x\right) = \frac{1}{4} {x}^{- \frac{3}{4}} = \frac{1}{4 {\sqrt{x}}^{3}}$

$f ' \left(81\right) = \frac{1}{4 {\sqrt{81}}^{3}} = \frac{1}{4 \cdot {3}^{3}} = \frac{1}{108}$

Therefore we can approximate (near $81$):

$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

$\implies \sqrt{x} \approx 3 + \frac{1}{108} \left(x - 81\right)$

So:

$\sqrt{84} = 3 + \frac{1}{108} \left(84 - 81\right)$

$3 + \frac{1}{108} \cdot 3 = \frac{324}{3} + \frac{3}{108} = \frac{327}{108} \approx 3.02778$

The more accurate value is $3.02740$

so the linear approximation is fairly close.

Feb 7, 2018

$\sqrt{84} \approx 3.02 \overline{7}$

#### Explanation:

We can say that we have a function of $f \left(x\right) = \sqrt{x}$
and $\sqrt{84} = f \left(84\right)$

Now, let's find the derivative of our function.

We use the power rule, which states that if $f \left(x\right) = {x}^{n}$, then $f ' \left(x\right) = n {x}^{n - 1}$ where $n$ is a constant.

$f \left(x\right) = {x}^{\frac{1}{4}}$

=>$f ' \left(x\right) = \frac{1}{4} \cdot {x}^{\frac{1}{4} - 1}$

=>$f ' \left(x\right) = \frac{{x}^{- \frac{3}{4}}}{4}$

=>$f ' \left(x\right) = \frac{1}{x} ^ \left(\frac{3}{4}\right) \cdot \frac{1}{4}$

=>$f ' \left(x\right) = \frac{1}{4 {x}^{\frac{3}{4}}}$

Now, to approximate $\sqrt{84}$, we try to find the perfect fourth-power closest to 84

Let's see...
$1$
$16$
$81$
$256$
.
.
.
We see that $81$ is our closest one.

We now find the tangent line of our function when $x = 81$

=>$f ' \left(81\right) = \frac{1}{4 \cdot {81}^{\frac{3}{4}}}$

=>$f ' \left(81\right) = \frac{1}{4 \cdot {81}^{\frac{2}{4}} \cdot {81}^{\frac{1}{4}}}$

=>$f ' \left(81\right) = \frac{1}{4 \cdot 9 \cdot 3}$

=>$f ' \left(81\right) = \frac{1}{108}$

This is the slope we are looking for.

Let's try to write the equation of the tangent line in the form $y = m x + b$

Well, what is $y$ equal to when $x = 81$?

Let's see...
$f \left(81\right) = \sqrt{81}$
=>$f \left(81\right) = 3$

Therefore, we now have:
$3 = m 81 + b$ We know that the slope, $m$, is $\frac{1}{108}$

=>$3 = \frac{1}{108} \cdot 81 + b$ We can now solve for $b$.

=>$3 = \frac{81}{108} + b$

=>$3 = \frac{3}{4} + b$

=>$2 \frac{1}{4} = b$

Therefore, the equation of the tangent line is $y = \frac{1}{108} x + 2 \frac{1}{4}$

We now use 84 in the place of $x$.

=>$y = \frac{1}{108} \cdot 84 + 2 \frac{1}{4}$

=>$y = \frac{1}{9} \cdot 7 + 2 \frac{1}{4}$

=>$y = \frac{7}{9} + \frac{9}{4}$

=>$y = \frac{28}{36} + \frac{81}{36}$

=>$y = \frac{109}{36}$

=>$y = 3.02 \overline{7}$

Therefore, $\sqrt{84} \approx 3.02 \overline{7}$