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Question
If f(x) = x3 + x – 2, find (f–1)'(0).
Solution
f(x) = x3 + x – 2 ...(1)
Differentiating w.r.t. x, we get
f'(x) = `"d"/"dx"(x^3 + x – 2)`
= 3x2 + 1 - 0
= 3x2 + 1
We know that
`(f^-1) ^' (y) = (1)/(f"'"(x)` ...(2)
From (1), y = f(x)
= -2, when x = 0
∴ from (2), (f-1)'(-2)
= `(1)/(f"'"(0)) = 1/(3x^2+1)`
`(1)/(f"'"(0)) = 1/(3(0)^2 + 1)` ....[∵ x = 0]
`(1)/(f"'"(0)) = 1/(0 + 1)`
`(1)/(f"'"(0)) = 1/1`
`(1)/(f"'"(0)) = 1`
∴ f'(0) = 1
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