Find Local Approximation of f(x)=x^(1/3) at x=26.6 with f(27)=3

[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, 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?

 

[teclo]

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?

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)

 

[UrbanXrisis]

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

 

[teclo]

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

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)