suppose you wanted to use linearization to estimate cuberoot 31, and then determine the error in that approximaition

Question

1. we need an equation with x and y, what is the equation
2. pick a point on the graph of the equation in part 1 that is close to the point that you are trying to estimate, what is that point
3. write the equation of the tangent line to the equation from part 1 using the point from part 2

Answer 1

Let f(x) = ³√x. Then f'(x) = (1/3)x^-2/3

If x is close to a, then f'(a) ≈ [f(x) - f(a)] / [x - a]

So, f(x) ≈ f(a) + f'(a)(x - a)

Let x = 31 and a = 27 (since we know the cube root of 27)

³√31 = f(31) ≈ f(27) + f'(27)(31 - 27) = 3 + (1/27)(4) = 3 4/27

Answer 2

Start with x = cuberoot(31) then cube both sides to get x³ = 31 then subtract 31 from both sides to get x³ - 31 = 0

Can you see that if we find the "zeros" of f(x) = x³ - 31, we would set x³ - 31 equal to zero? Therefore the function is f(x) = x³ - 31 (aka y = x³ - 31)

cuberoot(31) is close to cuberoot(27) which is 3 so my first estimate would be 3. If x = 3, then f(3) = -4 so the point is (3, -4)

Since the function is f(x) = x³ - 31, then (since the slope of the tangent line is the derivative), f '(x) = 3x² and at x = 3, the slope of the tangent line is f '(3) = 3(3²) = 27

That makes the equation of the tangent line: y - -4 = 27(x - 3) or y + 4 = 27x - 81 or y = 27x - 85