use a linear approximation to estimate the number (125.07)^2/3

Answer 1

We use a function f(x) = x^(2/3)

[I assume x ^ (2/3) ?, if you meant (x^2)/3, let me know]

We use linear approximation -> f'(x) = 2/3 * x ^ (-1/3) at x = 125

it is close enough to our x = 125.07

f'(125) = 2/3 * (125 ^ (-1/3)) = 2/3 * (1/5) = 2/15 = slope

Then we use tangent line. But, we need to have a point => f(125) = 5^2 = 25.

P(125,25) and m = 2/15

y - 25 = 2/15 * (x - 125)

Now at x = 125.07,

y -25 = 2/15 * (125.07 - 125)

y = 2/15 * .07 + 25 = 25.00933333333 ~~ 25.0093