How do you find the linearization of `f(x)=x^(3/4)` at x=1?

Here we need to locally substitute the actual curve of our function with the line tangent to our curve at point of coordinate `x=1`.

Basically instead of considering the real curve we substitute it with a straight line (the tangent). This works locally, i.e., in a very small interval around our point so we must be careful not to overdo it!

We will need the equation of the line in the general form:
`y-y_0=m(x-x_0)`
where:
`m` is the slope;
`x_0,y_0` are the coordinates of the point of interest.
In our case we have:
`x_0=1` (given)
substituting into our function we see that: `f(1)=y_0=1`
Now we need the slope `m`; we can find it by deriving our function and evaluating it at `x=1`:
`f'(x)=3/4x^(3/4-1)=3/4x^(-1/4)`
evaluated at `x=1`:
`f'(1)=m=3/4`

Now we use: `y-y_0=m(x-x_0)` to get:
`y-1=3/4(x-1)`
or
`y=3/4x-3/4+1`
`y=3/4x+1/4`

Graphically:

You can see that around `x=1` the line and the curve are almost the same thing, so, if you need, you can study the line instead of the original curve (where the curve may represent a phenomenon or tendency)!