Question

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

Answer 1

`root(4)(84) ~~ 3.03`

Explanation:

Note that `3^4 = 81`, which is close to `84`.

So `root(4)(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(x) = x^4-84`

Then:

`f'(x) = 4x^3`

and given an approximate zero `x=a` of `f(x)`, a better approximation is:

`a - (f(a))/(f'(a))`

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

`3-(f(3))/(f'(3)) = 3-(3^4-84)/(4(3)^3) = 3-(81-84)/(4 * 27) = 3+1/36 = 109/36 = 3.02bar(7)`

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

Answer 2

`root(4)(84)~~3.02778`

Explanation:

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

`f(x) ~~f(a)+f'(a)(x-a)`

If given: `f(x) = root(4)(x)`

then a suitable choice for `a` would be `a=81` because we know `root(4)81=3` exactly and it is close to `84`.

So:

`f(a) = f(81) = root(4)(81)=3`

Also;

`f(x) = x^(1/4)` so `f'(x) = 1/4x^(-3/4)=1/(4root(4)(x)^3)`

`f'(81) = 1/(4root(4)(81)^3)=1/(4*3^3)=1/108`

Therefore we can approximate (near `81`):

`f(x)~~f(a)+f'(a)(x-a)`

`implies root(4)(x)~~3+1/(108)(x-81)`

So:

`root(4)(84)=3+1/108(84-81)`

`3+1/108*3=324/3+3/108=327/108~~3.02778`

The more accurate value is `3.02740`

so the linear approximation is fairly close.

Answer 3

`root [4](84)~~3.02bar7`

Explanation:

We can say that we have a function of `f(x)= root(4) (x)`
and `root(4) (84)=f(84)`

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

We use the power rule, which states that if `f(x)=x^n`, then `f'(x)=nx^(n-1)` where `n` is a constant.

`f(x)=x^(1/4)`

=>`f'(x)=1/4*x^(1/4-1)`

=>`f'(x)=(x^(-3/4))/4`

=>`f'(x)=1/x^(3/4)*1/4`

=>`f'(x)=1/(4x^(3/4))`

Now, to approximate `root(4) (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'(81)=1/(4*81^(3/4))`

=>`f'(81)=1/(4*81^(2/4)*81^(1/4))`

=>`f'(81)=1/(4*9*3)`

=>`f'(81)=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=mx+b`

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

Let's see...
`f(81)=root(4)(81)`
=>`f(81)=3`

Therefore, we now have:
`3=m81+b` We know that the slope, `m`, is `1/108`

=>`3=1/108*81+b` We can now solve for `b`.

=>`3=81/108+b`

=>`3=3/4+b`

=>`2 1/4=b`

Therefore, the equation of the tangent line is `y=1/108x+2 1/4`

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

=>`y=1/108*84+2 1/4`

=>`y=1/9*7+2 1/4`

=>`y=7/9+9/4`

=>`y=28/36+81/36`

=>`y=109/36`

=>`y=3.02bar7`

Therefore, `root [4](84)~~3.02bar7`