- #1
UrbanXrisis
- 1,196
- 1
I need to find thelocal approximation of f(x)=x^(1/3) at x=26.6, knowing f(27)=3.
Here's what I did, don't know if I did it right:
f '(x)=(1/3)x^(-2/3)=[x^(-2/3)]/3
slope of the tanget = slope of the secant
[x^(-2/3)]/3=(y-y1)/(x-x1)
[26.6^(-2/3)]/3=(y-3)/(x-27)
now I sub in X and solve for y:
[26.6^(-2/3)]/3=(y-3)/(26.6-27)
y=2.985
I don't think that's the answer at all, could someone check it?
Here's what I did, don't know if I did it right:
f '(x)=(1/3)x^(-2/3)=[x^(-2/3)]/3
slope of the tanget = slope of the secant
[x^(-2/3)]/3=(y-y1)/(x-x1)
[26.6^(-2/3)]/3=(y-3)/(x-27)
now I sub in X and solve for y:
[26.6^(-2/3)]/3=(y-3)/(26.6-27)
y=2.985
I don't think that's the answer at all, could someone check it?