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Question
Find the approximate values of : f(x) = x3 – 3x + 5 at x = 1.99.
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Solution
f(x) = x3 – 3x + 5
∴ f'(x) = `d/dx(x^3 - 3x + 5)`
= 3x2 – 3 x 1 + 0
= 3x2 – 3
Take a = 2, h = – 0.01
Then f(a)
= f(2)
= (2)3 – 3(2) + 5
= 8 – 6 + 5
= 7
f'(a) = f'(2)
= 3(2)2 – 3
= 12 – 3
= 9
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ f(1.99) = f(2 – 0.01)
≑ f(2) – (0.01).f'(2)
≑ 7 – 0.01 x 9
= 7 – 0.09
= 6.91
∴ f(1.99) ≑ 6.91.
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