Linear Approximation

Question

Use linear approximation , i.e the tangent line, to approximate 3sqrt 8.02 as follow. Left f(x)=3sqrtx and find the equation of the tangent line to f(x) at x=8 in the fo9rm y=mx+b.

m =

b =

using these values , find the approximation

3sqrt8.02 =

Answer 1

If f(x) = 3√x, or f(x) = 3(x)^1/2, then f '(x) = (1/2)(3)(x)^-1/2 or f '(x) = 3/(2√x) so f'(8) = 3/(2√8) or 3/(4√2) or (3√2)/8

The derivative at x = 8, f '(8), represents the slope of the tangent line but we need to also know a point that is on the line.

So f(8) = 3√8 or 6√2 so the point is (8, 6√2)

Using the point-slope form:

y - 6√2 = [(3√2)/8](x - 8)

Now, multiply this out and combine like terms if any to get in the form of y = mx + b

y - 6√2 = [(3√2)/8]x - 3√2

y = [(3√2)/8]x + 3√2

so m = (3√2)/8 and b = 3√2

Now, to approximate 3√(8.02), you plug in x = 8.02 into y = [(3√2)/8]x + 3√2:

y ≈ [(3√2)/8](8.02) + 3√2

y ≈ 4.253247289 + 4.242640687

y ≈ 8.495887976