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

1 Answer
Jun 1, 2016

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

Explanation:

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]}