Use a linear approximation (or differentials) to estimate the given number. (Use the linearization of 1/x. Do?

Question

Use a linear approximation (or differentials) to estimate the given number. (Use the linearization of 1/x. Do not round the answer.)

1/99

Answer 1

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f(x) = 1/x ----> f '(x) = -1/x^2

1/99 is approximately 1/100 so close to this the gradient of the tangent is

-1/(100^2) = -0.0001

This means that for every unit to the right the function decreases by 0.0001 approximately and so for every unit to the left it increases by 0.0001 approximately.

Therefore 1/99 =approx= 1/100 + 0.0001 = 0.0101

Edit. A calculator will of course confirm that 1/99 = 0.010101010101.......

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Answer 2

metode 1)

f(x)=1/x

f '(x)= -1/x²

f(x + ∆x) ≈ f(x) + ∆x f '(x)

we have:

x= 100

∆x= -1

x + ∆x = 100 + (-1) = 99

f(100)= 1/100 = 0.01

f '(100) = -1/100² = -0.0001

hence

f(x + ∆x) ≈ f(x) + ∆x f '(x)

f(100 + (-1)) ≈ f(100) + (-1) f '(100)

f(99) ≈ 1/100 +(-1)(-1/100² )

f(99) ≈ 0.01 +(-1)(-0.0001)

1/99 ≈ 0.01 + 0.0001

1/99 ≈ 0.0101

metode 2)

f(x)=1/x

1/99 = 1/(100 - 1)

1/99 = (100 - 1)^(-1) : Binomial expansion

1/99 = 100^(-1) + (-1)(-1)100^(-2) + (-1)(-2)/2 (-1)^2 100^(-3) + . . .

1/99 = 1/100 + 1/100^(2) + 1/100^(3) + . . .

1/99 = 0.01 + 0.0001 + 0.000001 + . . .

1/99 = 0.0101010 . . .

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