How do you find the linearization at a=16 of #f(x)=x^(3/4)#?
1 Answer
Explanation:
First, find the point on the graph of
#f(16)=16^(3/4)=(2^4)^(3/4)=2^3=8#
To find the slope of the line, differentiate
The power rule shows us that the derivative of
#f'(x)=3/4x^(3/4-1)=3/4x^(-1/4)=3/(4x^(1/4))#
The slope of the linearization function is
#f'(16)=3/(4(16)^(1/4))=3/(4(2^4)^(1/4))=3/(4(2)^1)=3/8#
The linearization function
#L(x)=f(a)+f'(a)(x-a)#
Where
#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
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]}