How do you find the linearization at a=16 of `f(x)=x^(3/4)`?

First, find the point on the graph of `f(x)` when `x=16`.

`f(16)=16^(3/4)=(2^4)^(3/4)=2^3=8`

To find the slope of the line, differentiate `f(x)` through the power rule and then find the derivative's value at `x=16`.

The power rule shows us that the derivative of `f(x)=x^(3/4)` is

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

The slope of the linearization function is `f'(16)`:

`f'(16)=3/(4(16)^(1/4))=3/(4(2^4)^(1/4))=3/(4(2)^1)=3/8`

The linearization function `L(x)` at a point `x=a` can be written as:

`L(x)=f(a)+f'(a)(x-a)`

Where `a=16` this gives us:

`L(x)=f(16)+f'(16)(x-16)`

`L(x)=8+3/8(x-16)`

If you wish, this can be simplified into slope-intercept form:

`L(x)=3/8x+2`

As we zoom in on the function `f(x)=x^(3/4)` and the linearization function at `x=16`, we should see the two start more and more to resemble one another.

graph{(y-x^(3/4))(y-3/8x-2)=0 [-8.75, 64.32, -8.76, 27.78]}

graph{(y-x^(3/4))(y-3/8x-2)=0 [0.42, 40.97, -3.06, 17.23]}

graph{(y-x^(3/4))(y-3/8x-2)=0 [7.95, 27.95, 3.04, 13.05]}

graph{(y-x^(3/4))(y-3/8x-2)=0 [12.995, 19.925, 5.66, 9.128]}