Question

Let f(x) = `x + x^2/2 + x^3/3 + x^4/4 + x^5/5` and g(x) = f^–1(x), then |g''(0)| is ______.

Options

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

Let f(x) = `x + x^2/2 + x^3/3 + x^4/4 + x^5/5` and g(x) = f^–1(x), then |g''(0)| is 1.00.

Explanation:

f(x) = `x + x^2/2 + x^3/3 + x^4/4 + x^5/5`  (f(0) = 0)

f'(x) = 1 + x + x² + x³ + x⁴   (f'(0) = 1)

f''(x) = 1 + 2x + 3x² + 4x³  (f''(0) = 1)

Now f(g(x)) = x

Diff. w.r. to x

f'(g(x))g'(x) = 1  ...(i)

Again Diff. w.r. to x

f''(g(x))(g'(x))² + g''(x)f'(g(x)) = 0

x = 0

f''(g(0))(g'(0))² + g''(0)f'(g(0)) = 0  ...(ii)

as f(0) = 0 ⇒ g(0) = 0

and from (i) g'(0) = `1/(f^'(0))`

now from (ii)

`"f"^('')(0)(1/("f"^'(0)))^2 + "g"^('')(0)"f"^'(0)` = 0

1.1² + g''(0).1 = 0

g''(0) = –1

|g''(0)| = 1