# Find (f −1)'(a). f(x) = sqrt (2x^3 + 4x^2 + 2x + 1), a = 3?

### 3 Answers

- PopeLv 79 months ago
This one is in the same form as the one you asked nine hours ago, but there is this significant difference.

f(-1) = 1

f(0) = 1

This proves that function f is not injective. The function has no inverse, so the question has no answer.

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- AlanLv 79 months ago
Hopefully , you mean derivative of f inverse

where f(x) = sqrt( 2x^3 + 4x^2 + 2x + 1)

let u = 2x^3 + 4x^2 + 2x + 1

du/dx = 6x^2 + 8x + 2

f(u) = u^(1/2) =

f'(u) = (1/2) u^(-1/2)

f'(x) = du/dx * df/du = (6x^2 +8x + 2)* (1/2)/ sqrt(2x^3 + 4x^2 + 2x + 1)

f'(a) = (3a^2 + 4a + 1) /sqrt( 2a^3 + 4a^2 + 2a + 1)

f'(x) = df/dx

f_inverse'(x) = dx/df

f_inverse'(a) = sqrt(2a^3 + 4a^2 + 2a + 1)/ (3a^2 + 4a + 1)

but a =3

f_inverse'(3) = sqrt (2*3^3 + 4*3^2 + 2*3 + 1) / (3*3^2 + 4*3 + 1)

f_inverse'(3) = sqrt ( 2*27 + 4*9 + 6 + 1 ) / (27 + 12 + 1)

f_inverse'(3) = sqrt( 54 + 36 +7 ) / 40

f_inverse'(3) = sqrt(97)/ 40

f_inverse'(3) = approx. 0.246221445

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