# Use linear approximation i.e the tangent line ...?

I am very confused how to solve this, please help!

### 4 Answers

- ted sLv 74 weeks ago
y - f(a) = f ' ( a ) ( x - a ).......f ' = ( 1/4) x^(-3/4)...f '(81) = 1/108...a = 81& f(a) = 3

- Ian HLv 74 weeks ago
f(81) = 81^(1/4) = 3

f'(x) = (1/4)x^(-3/4)

f'(81) = (1/4)*81^(-3/4) = 1/(4*27) = 1/108

L(x) = f’(81)(x – 81) + f(81)

L(x) = (1/108)(x – 81) + 3

L(81.04) = 3 + (0.04/108) ~ 3.000370370370

Compare ... 81.04 ^(1/4) = 3.000370301803 calculator

- TomVLv 74 weeks ago
Let f(x) = x^(1/4)

f'(x) = (1/4)x^(-3/4)

For values of x near a:

f'(a) ≈ [f(x) - f(a)]/(x-a)

f(x) ≈ f(a) + f'(a)(x-a)

Let a = 81. Then f(81.04) = f(81) + f'(81)(0.04)

f(81) = 81^(1/4) = 3

f'(81) = (1/4)/27 = 1/108

f(81.04) = (81.04)^(1/4) ≈ 3 + 0.04/108 ≈ 3.000370370_

(Actual value, 3.0003703018... differs from the approximate value in the 8th decimal place)

- RealProLv 74 weeks ago
You are asked to find L(81.04). The tangent line approximates the function in a close neighborhood around the tangency point. I don't see how this can be confusing if you spent at least 5 minutes studying these things.