# Linear Approximation

1. Nov 9, 2004

### UrbanXrisis

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, dont 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 dont think that's the answer at all, could someone check it?

2. Nov 9, 2004

### teclo

your answer is reasonable. why? it's just under the cube root of 27, which is what you're looking at. cube that number to check your results -- 26.597 pretty close.

as i recall

the formula is L(x) = f(a) + f'(a)(x-a)

f(x) = x^(1/3)
f'(x) = 1/(3x^(2/3))

i arrive at the same answer plugging those values in.

f(x) = x^(1/3)

3. Nov 9, 2004

### UrbanXrisis

However, I'm not supposed to use a calculator to solve it...is there an easier way?

4. Nov 9, 2004

### teclo

well subsitute the values in fractional form

3 + 1/3(27)^2/3(26.6-27)

3+(1/27)(-4/10)

3-4/270 = 3-2/135 = 2.985

the fractional answer is correct, the decimal is simply an approximation of the fractional expressoin. (both are approximations, keep that in mind)