Use linear approximation, i.e the tangent line to approximate 3rd sqrt(1.1) as follows.?

Question

So approximate 3rd sqrt(1.1). Let f(x) = 3rd sqrt(x), the equation of the tangent line f(x)=1 can be written in the form y=mx+b.
Find m: Find b: And using this information approximate 3rd sqrt(1.1). Can anyone help me do this. I have found the slope to be 1/3 but I have no idea how to do the last two parts. Please Help.

thanks

erek

Answer 1

Use point-slope form:

y - y0 = m(x - x0)

where m=1/3, x0=1, and y0=f(1)=1. So you have y-1 = (1/3)(x-1).

Solve for y to obtain the equation in y=mx+b form:

y = 1 + x/3 - 1/3 = x/3 + 2/3.

Now plug in x=1.1, and you get y= (1.1)/3 + 2/3 = 1.0333333 (repeating)

Answer 2

k
1. take the derivative of y=cube root of (x)
y'=1/3 x^{-2/3}
now,
if x=1
then
y'=1/3
and the eq. of the tangetn line is:
y-1=1/3 (x-1)
so y=1/3 x + 2/3
now for the aprox
simply substitute x=1.1
so you get
y= 1/3(1.1) +2/3