Find the linear approximation of the function f(x)=^3sqrt(1+3x) at a=0 and use it to approximate ^3sqrt(1.03)Is this an overestimate or underestimate of the actual value? Explain your answer in terms of the tangent line and the curve.

Answer 1

if ^3sqrt means cube root, then
f(x) = (1+3x)^(1/3)
df/dx = (1+3x)^(-2/3)
so, df = (1+3x)^(-2/3) dx
At x=0, df = dx
So, since dx=1.03, f=f(0)+1.03=1.03
at x=0, the line y=x+1 is tangent to the curve

I think you meant to approximate by using x=1. In that case,
f(1) = ∛4 = 1.587
df = 1/∛16 dx
since dx=.03, df=.03/∛16 = .015/∛2 = .0119
f(1.03) = 1.587+0.0119 = 1.599
at x=1, the tangent line is

y-1.587 = .3969(x-1) or
y = .3969x + 1.19

Visit http://rechneronline.de/function-graphs

and enter the three functions as
(1+3x)^(1/3)
x+1
.3969x + 1.19
then click "Draw" and you will see how the lines approximate the curve.

Answer 2

at x=0, since f(0) = 1, the
approximation at x=1.03 would be
f(1.03) = 1+1.03 = 2.03

Answer 3

In f(1.03) = 1.587+0.0119 = 1.599, did you apply the linear approximation equation L(x)=f(a)+f'(a)(x-a)? Because I'm still a little confused on how to apply that equation to problems like these.

Answer 4

That's exactly what I did.
f(1) = 1.587
f'(1) = .3968
(1.03-1) = .03