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

1 Answer
Aug 1, 2016

I found: #y=3/4x+1/4#

Explanation:

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:
enter image source here

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)!